Work Done by Constant Force Questions in English

(A) The displacement vector is $\vec{s} = -3\hat{i}\, m$.
The forces acting on the trunk are:
$\vec{F}_1 = -5\hat{i}\, N$
$\vec{F}_2 = 9\cos(60^\circ)\hat{i} + 9\sin(60^\circ)\hat{j} = 4.5\hat{i} + 4.5\sqrt{3}\hat{j}\, N$
$\vec{F}_3 = -3\hat{j}\, N$
The net force $\vec{F}_{net}$ is:
$\vec{F}_{net} = \vec{F}_1 + \vec{F}_2 + \vec{F}_3 = (-5 + 4.5)\hat{i} + (4.5\sqrt{3} - 3)\hat{j} = -0.5\hat{i} + (4.5\sqrt{3} - 3)\hat{j}\, N$
The net work done $W$ is the dot product of the net force and displacement:
$W = \vec{F}_{net} \cdot \vec{s} = (-0.5\hat{i} + (4.5\sqrt{3} - 3)\hat{j}) \cdot (-3\hat{i})$
$W = (-0.5 \times -3) + (0) = 1.5\, J$.

(A) The work done on the cycle by the road is the work done by the stopping (frictional) force on the cycle due to the road.
$(a)$ The stopping force and the displacement make an angle of $180^{\circ} \; (\pi \; \text{rad})$ with each other. Thus,the work done by the road,$W_{r} = F d \cos \theta$
$W_{r} = 200 \times 10 \times \cos(180^{\circ})$
$W_{r} = 2000 \times (-1) = -2000 \; J$
It is this negative work that brings the cycle to a halt in accordance with the Work-Energy theorem.
$(b)$ From Newton's Third Law,an equal and opposite force acts on the road due to the cycle. Its magnitude is $200 \; N$. However,the road undergoes no displacement $(d = 0)$. Thus,the work done by the cycle on the road is $W = F \times 0 = 0 \; J$.

(D) The force acting on the body is given by $F = -\hat{i} + 2\hat{j} + 3\hat{k} \text{ N}$.
The displacement of the body is along the $z$-axis for a distance of $4 \text{ m}$,so the displacement vector is $s = 4\hat{k} \text{ m}$.
Work done $(W)$ is defined as the dot product of force and displacement: $W = F \cdot s$.
$W = (-\hat{i} + 2\hat{j} + 3\hat{k}) \cdot (4\hat{k})$.
Using the properties of unit vectors $(\hat{i} \cdot \hat{k} = 0, \hat{j} \cdot \hat{k} = 0, \hat{k} \cdot \hat{k} = 1)$:
$W = (-1 \times 0) + (2 \times 0) + (3 \times 4) = 0 + 0 + 12 = 12 \text{ J}$.
Therefore,the work done by the force is $12 \text{ J}$.

(N/A) Work: The product of the magnitude of force and the displacement in the direction of the force is called work.
$\therefore$ Work $=$ Force $\times$ Displacement
If the force and the displacement are in the same direction,the work done is positive. If they are in opposite directions,the work done is negative. If the force is perpendicular to the displacement,the work done is zero.
Kinetic Energy: The energy possessed by a body by virtue of its motion is known as kinetic energy.
Alternatively,kinetic energy is defined as half of the product of the mass of a body and the square of its velocity.
$\therefore$ Kinetic energy $= \frac{1}{2} m v^{2}$
where $m$ is the mass of the body and $v$ is the velocity of the body.

(N/A) Constant force: When the magnitude and direction of a force acting on a body remain constant during its motion,such a force is known as a constant force.
Work: The work done by a constant force is defined as the dot product of the force vector $\vec{F}$ and the displacement vector $\vec{d}$.
$\therefore W = \vec{F} \cdot \vec{d} = F d \cos \theta$
where $F$ is the magnitude of the force,$d$ is the magnitude of displacement,and $\theta$ is the angle between the force vector and the displacement vector.
If the displacement is not in the direction of the applied force,work can be calculated in two ways:
$(1)$ Work $=$ (Force) $\times$ (Component of displacement in the direction of force) $= F(d \cos \theta)$
$(2)$ Work $=$ (Component of force in the direction of displacement) $\times$ (Displacement) $= (F \cos \theta)d$
As shown in the figure,when a force $\vec{F}$ acts at an angle $\theta$ to the displacement $\vec{d}$:
$(i)$ The component of force in the direction of displacement is $F \cos \theta$,which is responsible for the work done.
$(ii)$ The perpendicular component $F \sin \theta$ does no work because it is perpendicular to the displacement.

(N/A) The general formula for work done is $W = \vec{F} \cdot \vec{d} = F d \cos \theta$,where $\theta$ is the angle between the force vector $\vec{F}$ and the displacement vector $\vec{d}$.
$1$. Positive Work: If the angle $\theta$ between the force and displacement is in the range $0^{\circ} \le \theta < 90^{\circ}$,then $\cos \theta$ is positive. Consequently,the work done $W$ is positive. In this case,the force aids the motion of the body.
$2$. Negative Work: If the angle $\theta$ between the force and displacement is in the range $90^{\circ} < \theta \le 180^{\circ}$,then $\cos \theta$ is negative. Consequently,the work done $W$ is negative. In this case,the force opposes the motion of the body.

(N/A) Work is defined as the product of the magnitude of the force applied to an object and the displacement of the object in the direction of the force.
Mathematically,it is expressed as the dot product of the force vector $\vec{F}$ and the displacement vector $\vec{d}$:
$W = \vec{F} \cdot \vec{d} = Fd \cos \theta$
Where:
$W$ is the work done,
$F$ is the magnitude of the force,
$d$ is the magnitude of the displacement,
$\theta$ is the angle between the force vector and the displacement vector.
Work is a scalar quantity and its $SI$ unit is the Joule $(J)$.

(C) force is defined as a vector quantity,meaning it has both magnitude and direction. For a force to be considered constant,both its magnitude and its direction must remain unchanged throughout the entire process of displacement. If either the magnitude or the direction changes,the force is considered a variable force. Therefore,the correct condition is that both the magnitude and the direction of the force remain constant.

(B) Work done by a body against a force means the displacement of the body is in the direction opposite to the applied force.
Since the formula for work is $W = F \cdot s \cdot \cos(\theta)$,where $\theta$ is the angle between the force and displacement.
When the body moves against the force,the angle $\theta = 180^{\circ}$.
Therefore,$W = F \cdot s \cdot \cos(180^{\circ}) = F \cdot s \cdot (-1) = -F \cdot s$.
Thus,the work done is negative.

(B) The work done by a force is given by the formula $W = F d \cos \theta$,where $F$ is the force,$d$ is the displacement,and $\theta$ is the angle between the force and displacement vectors.
Here,the frictional force $F = 200 \, N$ acts in the direction opposite to the displacement $d = 10 \, m$.
Therefore,the angle $\theta = 180^{\circ}$ (or $\pi$ radians).
Substituting these values: $W = 200 \times 10 \times \cos(180^{\circ})$.
Since $\cos(180^{\circ}) = -1$,we get $W = 2000 \times (-1) = -2000 \, J$.

(A) The work done is positive.
This is because the displacement of a freely falling object occurs in the same direction as the gravitational force acting on it.
According to the formula for work,$W = F \cdot d \cdot \cos(\theta)$,where $\theta$ is the angle between the force and displacement.
Since both the gravitational force and the displacement are directed downwards,$\theta = 0^\circ$.
Therefore,$W = F \cdot d \cdot \cos(0^\circ) = F \cdot d$,which is positive.

(N/A) The work done by a force $\overrightarrow{F}$ over a displacement $d\overrightarrow{l}$ is given by $dW = \overrightarrow{F} \cdot d\overrightarrow{l}$.
Given that the work done is zero,we have $\overrightarrow{F} \cdot d\overrightarrow{l} = 0$.
Since the velocity is defined as $\vec{v} = \frac{d\overrightarrow{l}}{dt}$,we can write $d\overrightarrow{l} = \vec{v} dt$.
Substituting this into the work equation: $\overrightarrow{F} \cdot (\vec{v} dt) = 0$.
Since $dt \neq 0$,it follows that $\overrightarrow{F} \cdot \vec{v} = 0$.
This implies that the force $\overrightarrow{F}$ must always be perpendicular to the velocity vector $\vec{v}$ of the particle.
If the particle changes its direction of motion,the velocity vector $\vec{v}$ changes.
To maintain the condition $\overrightarrow{F} \cdot \vec{v} = 0$ (i.e.,the angle $\theta$ between $\overrightarrow{F}$ and $\vec{v}$ remains $90^{\circ}$),the direction of the force $\overrightarrow{F}$ must also change in accordance with the change in the direction of $\vec{v}$.
Therefore,the force must be velocity-dependent.

(B) For work to be done on an object when a force is applied,the object must undergo a displacement in a direction that is not perpendicular to the direction of the applied force. Mathematically,work is defined as $W = \vec{F} \cdot \vec{d} = Fd cos \theta$. For $W \neq 0$,the displacement $\vec{d}$ must have a component along the force $\vec{F}$,meaning $\theta \neq 90^{\circ}$.

(C) Work done $(W)$ is defined by the formula $W = F \cdot d \cdot \cos(\theta)$,where $F$ is the applied force,$d$ is the displacement,and $\theta$ is the angle between the force and displacement vectors.
Since the force $(F)$ and the displacement $(d)$ are the same for both objects,and assuming the force is applied in the direction of displacement $(\theta = 0^\circ)$,the work done is $W = F \cdot d$.
Because $F$ and $d$ are identical for both,the work done on both objects is equal,regardless of their masses.

(B) The work done $W$ by a constant force $\vec{F}$ during a displacement $\vec{d}$ is given by the dot product: $W = \vec{F} \cdot \vec{d}$.
Given the force vector $\vec{F} = (-1, 2, 3) \,N$.
The displacement is along the $X$-axis by $4 \,m$,so the displacement vector is $\vec{d} = (4, 0, 0) \,m$.
Calculating the dot product:
$W = (-1 \times 4) + (2 \times 0) + (3 \times 0)$
$W = -4 + 0 + 0 = -4 \,J$.
Therefore,the work done by the force is $-4 \,J$.

(N/A) Force is applied on the body to lift it in the upward direction,and the displacement of the body is also in the upward direction.
Since the angle between the applied force and displacement is $0^{\circ}$,the work done by the applied force is $W = F d \cos 0^{\circ} = F d$,which is positive.
$(b)$ The gravitational force acts in the downward direction,while the displacement is in the upward direction.
Since the angle between the gravitational force and displacement is $180^{\circ}$,the work done by the gravitational force is $W = F d \cos 180^{\circ} = -F d$,which is negative.

(A) The force of gravity acts on the car vertically downward,while the car is moving along a horizontal road. Therefore,the angle between the force of gravity and the displacement is $\theta = 90^{\circ}$.
The formula for work done is $W = F d \cos \theta$.
Substituting the values,we get $W = F d \cos 90^{\circ}$.
Since $\cos 90^{\circ} = 0$,the work done is $W = 0\,J$.
Thus,the work done by the car against gravity is $0\,J$ because the displacement is perpendicular to the gravitational force.

(N/A) Given: Weight of adult $W = mg = 600 \, N$,height raised per step $h = 0.25 \, m$,step length $l = 1 \, m$,total distance $d = 6 \, km = 6000 \, m$.
Total number of steps $n = \frac{d}{l} = \frac{6000 \, m}{1 \, m} = 6000$.
Energy utilised per step $E_{step} = mgh = 600 \, N \times 0.25 \, m = 150 \, J$.
Total energy utilised in jogging $E_{total} = n \times E_{step} = 6000 \times 150 \, J = 9 \times 10^5 \, J$.
Given that the body converts $10 \, \%$ of food energy intake into work,let $E_{food}$ be the total energy intake.
$0.10 \times E_{food} = 9 \times 10^5 \, J$.
$E_{food} = \frac{9 \times 10^5}{0.10} = 9 \times 10^6 \, J$.

(A) According to the work-energy theorem,the total work done on the block is equal to the change in its kinetic energy.
$W_{\text{total}} = \Delta K$
$W_F + W_{\text{spring}} = K_f - K_i$
Here,the work done by the constant force $F$ is $W_F = Fx$.
The work done by the spring force is $W_{\text{spring}} = -\frac{1}{2} k x^2$.
The initial kinetic energy $K_i = 0$ (as it starts from rest) and the final kinetic energy $K_f = \frac{1}{2} m v^2$.
Substituting these values into the equation:
$Fx - \frac{1}{2} k x^2 = \frac{1}{2} m v^2 - 0$
$Fx - \frac{1}{2} k x^2 = \frac{1}{2} m v^2$
Multiplying both sides by $2$:
$2Fx - k x^2 = m v^2$
$v^2 = \frac{2Fx - k x^2}{m}$
$v = \sqrt{\frac{2Fx - k x^2}{m}}$

(A) Given that the work done by both persons is equal: $W_A = W_B$.
The formula for work done is $W = Fd \cos \theta$.
For person $A$: $W_A = F_A d \cos 45^{\circ}$.
For person $B$: $W_B = F_B d \cos 60^{\circ}$.
Equating the two: $F_A d \cos 45^{\circ} = F_B d \cos 60^{\circ}$.
Substituting the values of $\cos 45^{\circ} = \frac{1}{\sqrt{2}}$ and $\cos 60^{\circ} = \frac{1}{2}$:
$F_A \times \frac{1}{\sqrt{2}} = F_B \times \frac{1}{2}$.
Rearranging for the ratio $\frac{F_A}{F_B}$:
$\frac{F_A}{F_B} = \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{(\sqrt{2})^2} = \frac{1}{\sqrt{2}}$.
Comparing this with $\frac{1}{\sqrt{x}}$,we get $x = 2$.

(D) According to the Work-Energy Theorem,the net work done on an object is equal to the change in its kinetic energy.
Since the suitcase is moved with a constant velocity,the change in kinetic energy $\Delta K.E. = 0$.
Therefore,the total work done by all forces is $W_{\text{Porter}} + W_{\text{gravity}} = 0$.
This implies $W_{\text{Porter}} = -W_{\text{gravity}}$.
The work done by gravity is $W_{\text{gravity}} = mgh$,where $h$ is the displacement.
Here,$m = 80\, \text{kg}$,$g = 9.8\, \text{m/s}^2$,and $h = 80\, \text{cm} = 0.8\, \text{m}$.
Since the porter is lowering the suitcase,the force applied by the porter is upwards while the displacement is downwards.
$W_{\text{Porter}} = -mgh = -(80) \times (9.8) \times (0.8) = -627.2\, \text{J}$.

(A) The correct option is $A$.
Since the force $\vec{F}$ is perpendicular to the velocity $\vec{V}$ of the particle,the component of the force in the direction of motion is zero $(F \cos 90^{\circ} = 0)$.
According to Newton's second law,the rate of change of speed depends on the tangential component of the force. Since this component is zero,the speed of the particle remains constant.
However,the force causes a change in the direction of the velocity vector,meaning the velocity is not uniform. Consequently,since momentum $\vec{p} = m\vec{v}$,the momentum also changes because the direction of the velocity changes.

(D) The force $\vec{F}$ is a constant force given by $\vec{F} = (3 \hat{i} + 4 \hat{j}) \;N$.
The particle moves from the origin $(0,0) \;m$ to $(4,0) \;m$ and then to $(4,3) \;m$. The total displacement vector $\vec{d}$ is the vector from the initial position $(0,0)$ to the final position $(4,3)$,which is $\vec{d} = (4 \hat{i} + 3 \hat{j}) \;m$.
Since the force is constant,the work done $W$ is given by the dot product of the force and the displacement vector:
$W = \vec{F} \cdot \vec{d}$
$W = (3 \hat{i} + 4 \hat{j}) \cdot (4 \hat{i} + 3 \hat{j})$
$W = (3 \times 4) + (4 \times 3)$
$W = 12 + 12 = 24 \;J$.
Therefore,the total work done is $24 \;J$.

(B) The maximum height reached by the projectile is $H = \frac{u^2 \sin^2 \theta}{2g}$.
During the upward motion,the displacement of the particle is in the upward direction,while the gravitational force $(F_g = mg)$ acts in the downward direction.
Therefore,the work done by gravity is $W = F_g \cdot H \cdot \cos(180^{\circ})$.
$W = -mg \times \left( \frac{u^2 \sin^2 \theta}{2g} \right)$.
$W = -\frac{m u^2 \sin^2 \theta}{2}$.

(D) According to Newton's second law of motion,the net force on the block is $F_{net} = T - mg = ma$.
Given that the acceleration $a = \frac{g}{3}$,we can substitute this into the equation:
$T - mg = m(\frac{g}{3})$
$T = mg + \frac{mg}{3} = \frac{4}{3} mg$
Since the tension $T$ acts in the same direction as the displacement $h$,the work done by the tension is given by $W = T \cdot h$.
$W = (\frac{4}{3} mg) \cdot h = \frac{4}{3} mgh$.

(C) The work done by a force is given by the formula $W = F \cdot s \cdot \cos(\theta)$,where $F$ is the force applied,$s$ is the displacement,and $\theta$ is the angle between the force and the displacement vector.
When a man carries a load on his head,he must apply an upward force equal to the weight of the load $(F = mg)$ to counteract gravity.
If he moves the load horizontally,the displacement is perpendicular to the force of gravity (the force he applies to hold the load),so $\theta = 90^\circ$. Since $\cos(90^\circ) = 0$,the work done against gravity is $0$.
If he lifts the load vertically upwards,the displacement is in the same direction as the force applied,so $\theta = 0^\circ$. Since $\cos(0^\circ) = 1$,the work done is $W = F \cdot s = mgs$.
Therefore,the maximum work is done when he lifts the load vertically upwards.

(C) The work done by a constant force is given by the dot product of force and displacement: $W = \vec{F} \cdot \vec{d} = F d \cos \theta$.
Since the force $F$ is constant and horizontal,the work done is $W = F \times (\text{horizontal displacement})$.
From the geometry of the circular arc,the horizontal displacement of the block as it moves from $A$ to $B$ is $d = R \sin 60^{\circ}$.
Substituting the value of $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$,we get the horizontal displacement $d = R \left(\frac{\sqrt{3}}{2}\right)$.
Therefore,the work done is $W = F \times \left(R \frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3}}{2} F R$.
Thus,the correct option is $C$.

(C) According to Newton's second law of motion,the net force acting on the block is $F_{net} = ma$.
Since the block is being pulled vertically upwards,the forces acting on it are the tension $T$ (upwards) and the weight $mg$ (downwards).
Therefore,$T - mg = ma$.
Given that the acceleration $a = \frac{g}{4}$,we substitute this into the equation:
$T - mg = m\left(\frac{g}{4}\right)$
$T = mg + \frac{mg}{4} = \frac{5}{4}mg$.
The work done by the tension $T$ is given by $W = T \cdot h \cdot \cos(0^\circ)$,since the displacement is in the same direction as the tension.
$W = \left(\frac{5}{4}mg\right) \cdot h = \frac{5}{4}mgh$.
Thus,the correct option is $C$.

(D) According to the Work-Energy Theorem,the work done by the net force on an object is equal to the change in its kinetic energy.
$W = \Delta K = K_f - K_i$
Since the object starts from rest,the initial kinetic energy $K_i = 0$.
The work done by a constant force $F$ over a distance $d$ is $W = F \cdot d$.
Therefore,$K_f = F \cdot d$.
Since $F$ and $d$ are constants,the kinetic energy $K_f$ is independent of the mass $m$ of the object.
Thus,the correct option is $D$.

(B) The work done $W$ by a constant force $\vec{F}$ is given by the dot product of the force and the displacement vector $\vec{d} = \vec{r}_f - \vec{r}_i$.
Given:
Force $\vec{F} = 5 \hat{i} + 2 \hat{j} + 7 \hat{k} \text{ N}$
Initial position $\vec{r}_i = 2 \hat{i} + 3 \hat{j} - 4 \hat{k}$
Final position $\vec{r}_f = 5 \hat{i} - 2 \hat{j} + \hat{k}$
Displacement $\vec{d} = \vec{r}_f - \vec{r}_i = (5-2) \hat{i} + (-2-3) \hat{j} + (1 - (-4)) \hat{k} = 3 \hat{i} - 5 \hat{j} + 5 \hat{k}$
Work done $W = \vec{F} \cdot \vec{d} = (5 \hat{i} + 2 \hat{j} + 7 \hat{k}) \cdot (3 \hat{i} - 5 \hat{j} + 5 \hat{k})$
$W = (5 \times 3) + (2 \times -5) + (7 \times 5)$
$W = 15 - 10 + 35 = 40 \text{ J}$.

(A) The work done against gravity is given by the change in gravitational potential energy,which is $W = mgh$,where $m$ is the mass,$g$ is the acceleration due to gravity,and $h$ is the vertical displacement.
In both cases,the mass $m = 50 \ kg$ and the vertical height $h = 20 \ m$ are the same.
Since gravity is a conservative force,the work done against it depends only on the initial and final vertical positions,not on the path taken.
Therefore,the work done in Case-$1$ is $W_1 = mgh = 50 \times g \times 20 = 1000g \ J$.
The work done in Case-$2$ is also $W_2 = mgh = 50 \times g \times 20 = 1000g \ J$.
The ratio of work done is $W_1 : W_2 = 1000g : 1000g = 1: 1$.
Thus,option $A$ is correct.

(B) Work done $(W)$ is defined as the dot product of force $(\vec{F})$ and displacement $(\vec{S})$: $W = \vec{F} \cdot \vec{S}$.
Given $\vec{F} = 2\hat{i} + b\hat{j} + \hat{k}$ and $\vec{S} = \hat{i} - 2\hat{j} - \hat{k}$.
Since the work done is zero,$W = (2\hat{i} + b\hat{j} + \hat{k}) \cdot (\hat{i} - 2\hat{j} - \hat{k}) = 0$.
Calculating the dot product: $(2)(1) + (b)(-2) + (1)(-1) = 0$.
$2 - 2b - 1 = 0$.
$1 - 2b = 0$.
$2b = 1$.
Therefore,$b = \frac{1}{2}$.

(A) The mass of the hanging part is $m' = \frac{m}{\ell} \times \frac{2 \ell}{5} = \frac{2m}{5}$.
The center of mass of the hanging part is at a distance $h = \frac{1}{2} \times \frac{2 \ell}{5} = \frac{\ell}{5}$ below the edge of the table.
The work done to pull the hanging part onto the table is equal to the change in potential energy of the hanging part.
$W = m'gh = \left( \frac{2m}{5} \right) g \left( \frac{\ell}{5} \right) = \frac{2mg \ell}{25}$.

(B) The work done $W$ by a force $\vec{F}$ during a displacement $\vec{s}$ is given by the dot product: $W = \vec{F} \cdot \vec{s}$.
Given,$\vec{F} = -5 \hat{i} - 7 \hat{j} + 3 \hat{k}$ and $\vec{s} = 3 \hat{i} - 2 \hat{j} + a \hat{k}$.
Substituting the values into the formula:
$14 = (-5 \hat{i} - 7 \hat{j} + 3 \hat{k}) \cdot (3 \hat{i} - 2 \hat{j} + a \hat{k})$
$14 = (-5)(3) + (-7)(-2) + (3)(a)$
$14 = -15 + 14 + 3a$
$14 = -1 + 3a$
$15 = 3a$
$a = 5$.

(C) According to the Work-Energy Theorem,the work done by the net force on an object is equal to the change in its kinetic energy.
$W = \Delta K.E.$
Since the body starts from rest,the initial kinetic energy is $0$.
$W = K.E._{final} - 0 = K.E._{final}$
The work done by a constant force $F$ over a distance $d$ is given by $W = F \cdot d$.
Therefore,$K.E. = F \cdot d$.
Since the force $F$ and the distance $d$ are constants,the kinetic energy acquired is independent of the mass $M$ of the body.

(D) The displacement vector $\vec{d}$ is given by $\vec{d} = \vec{r}_Q - \vec{r}_P$.
Given $\vec{r}_P = (3 \hat{i} + 4 \hat{j} + 5 \hat{k}) \ m$ and $\vec{r}_Q = (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) \ m$.
So,$\vec{d} = (2-3) \hat{i} + (3-4) \hat{j} + (4-5) \hat{k} = (-1 \hat{i} - 1 \hat{j} - 1 \hat{k}) \ m$.
The work done $W$ by a constant force $\vec{F}$ is given by the dot product $W = \vec{F} \cdot \vec{d}$.
$W = (3 \hat{i} + 4 \hat{j} + 5 \hat{k}) \cdot (-1 \hat{i} - 1 \hat{j} - 1 \hat{k})$.
$W = (3 \times -1) + (4 \times -1) + (5 \times -1) = -3 - 4 - 5 = -12 \ J$.

(D) The work done $W$ by a constant force $F$ is given by the formula: $W = F \cdot S \cdot \cos(\theta)$, where $S$ is the displacement and $\theta$ is the angle between the force and the displacement.
Given:
Force $F = 30 \,kg-wt = 30 \times 9.8 \,N = 294 \,N$
Displacement $S = 20 \,m$
Angle $\theta = 60^{\circ}$
Substituting these values into the formula:
$W = 294 \times 20 \times \cos(60^{\circ})$
Since $\cos(60^{\circ}) = 0.5$, we get:
$W = 294 \times 20 \times 0.5$
$W = 294 \times 10 = 2940 \,J$
Therefore, the work done by the gardener is $2940 \,J$.

(C) The work done by a force is defined as $W = \vec{F} \cdot \vec{s} = Fs \cos \theta$,where $\theta$ is the angle between the force and the displacement.
In this case,the normal reaction force $N$ acts vertically upwards,perpendicular to the horizontal surface.
The displacement $s$ of the block is along the horizontal surface.
Therefore,the angle $\theta$ between the normal reaction $N$ and the displacement $s$ is $90^\circ$.
Since $\cos 90^\circ = 0$,the work done by the normal reaction is $W = Ns \cos 90^\circ = 0$.

(C) The work done $W$ by a constant force $\vec{F}$ is given by the dot product of force and displacement: $W = \vec{F} \cdot \vec{d}$.
First,calculate the displacement vector $\vec{d} = \vec{r_2} - \vec{r_1}$.
$\vec{d} = (6 - 3) \hat{\imath} + (-1 - 2) \hat{\jmath} + (4 - (-1)) \hat{k} = 3 \hat{\imath} - 3 \hat{\jmath} + 5 \hat{k} \text{ m}$.
Now,calculate the work done:
$W = (5 \hat{\imath} - 2 \hat{\jmath} + 3 \hat{k}) \cdot (3 \hat{\imath} - 3 \hat{\jmath} + 5 \hat{k})$.
$W = (5 \times 3) + (-2 \times -3) + (3 \times 5) = 15 + 6 + 15 = 36 \text{ J}$.

(B) We know that the work done $W$ is given by the dot product of force $\vec{F}$ and displacement $\vec{d}$.
Given,force $\vec{F} = 2 \hat{i} - \hat{j} - \hat{k}$.
The object moves from the origin $(0, 0, 0)$ to the position vector $\vec{r} = 3 \hat{i} + 2 \hat{j} - 5 \hat{k}$,so the displacement $\vec{d} = \vec{r} - 0 = 3 \hat{i} + 2 \hat{j} - 5 \hat{k}$.
Using the property of dot product $\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1$ and $\hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 0$,we calculate:
$W = \vec{F} \cdot \vec{d} = (2 \hat{i} - \hat{j} - \hat{k}) \cdot (3 \hat{i} + 2 \hat{j} - 5 \hat{k})$
$W = (2 \times 3) + (-1 \times 2) + (-1 \times -5)$
$W = 6 - 2 + 5 = 9 \text{ units}$.

(D) The work done $W$ by a constant force $\vec{F}$ is given by the dot product of the force vector and the displacement vector $\vec{d}$.
Given $\vec{F} = (3 \hat{i} - 2 \hat{j}) \text{ N}$.
The initial position vector is $\vec{r}_1 = (1 \hat{i} + 2 \hat{j}) \text{ m}$.
The final position vector is $\vec{r}_2 = (2 \hat{i} + 0 \hat{j}) \text{ m}$.
The displacement vector $\vec{d} = \vec{r}_2 - \vec{r}_1 = (2-1) \hat{i} + (0-2) \hat{j} = (1 \hat{i} - 2 \hat{j}) \text{ m}$.
Work done $W = \vec{F} \cdot \vec{d} = (3 \hat{i} - 2 \hat{j}) \cdot (1 \hat{i} - 2 \hat{j})$.
$W = (3 \times 1) + (-2 \times -2) = 3 + 4 = 7 \text{ J}$.

(A) The work done $W$ by a constant force $\vec{F}$ is given by the dot product of the force vector and the displacement vector $\vec{d}$.
Given force $\vec{F} = (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) \text{ N}$.
The initial position vector is $\vec{r}_1 = (3 \hat{i} - 4 \hat{k}) \text{ m}$.
The final position vector is $\vec{r}_2 = (2 \hat{i} + 2 \hat{j} + 3 \hat{k}) \text{ m}$.
The displacement vector $\vec{d} = \vec{r}_2 - \vec{r}_1 = (2 - 3) \hat{i} + (2 - 0) \hat{j} + (3 - (-4)) \hat{k} = (-1 \hat{i} + 2 \hat{j} + 7 \hat{k}) \text{ m}$.
Work done $W = \vec{F} \cdot \vec{d} = (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) \cdot (-1 \hat{i} + 2 \hat{j} + 7 \hat{k})$.
$W = (2 \times -1) + (3 \times 2) + (4 \times 7) = -2 + 6 + 28 = 32 \text{ J}$.

(C) The work done $W$ by a constant force $\vec{F}$ during a displacement $\vec{d}$ is given by the dot product of the force and displacement vectors:
$W = \vec{F} \cdot \vec{d}$
Given:
$\vec{F} = (3 \hat{i} + 2 \hat{j} + 5 \hat{k}) \text{ N}$
$\vec{d} = (2 \hat{i} + 2 \hat{j} + 1 \hat{k}) \text{ m}$
Calculating the dot product:
$W = (3 \hat{i} + 2 \hat{j} + 5 \hat{k}) \cdot (2 \hat{i} + 2 \hat{j} + 1 \hat{k})$
$W = (3 \times 2) + (2 \times 2) + (5 \times 1)$
$W = 6 + 4 + 5$
$W = 15 \text{ J}$
Therefore,the work done is $15 \text{ J}$.

(B) Given force $\vec{F} = (4 \hat{i} + 2 \hat{j} + \hat{k}) \text{ N}$.
Initial position $\vec{r_1} = (2 \hat{i} + 2 \hat{j} + \hat{k}) \text{ m}$.
Final position $\vec{r_2} = (4 \hat{i} + 3 \hat{j} + 2 \hat{k}) \text{ m}$.
Displacement $\vec{S} = \vec{r_2} - \vec{r_1} = (4-2) \hat{i} + (3-2) \hat{j} + (2-1) \hat{k} = (2 \hat{i} + \hat{j} + \hat{k}) \text{ m}$.
Work done $W = \vec{F} \cdot \vec{S}$.
$W = (4 \hat{i} + 2 \hat{j} + \hat{k}) \cdot (2 \hat{i} + \hat{j} + \hat{k})$.
$W = (4 \times 2) + (2 \times 1) + (1 \times 1) = 8 + 2 + 1 = 11 \text{ J}$.

(A) Given:
Mass of the block,$m = 2 \,kg$
Angle of inclination,$\theta = 30^{\circ}$
Distance moved,$d = 4 \,m$
Acceleration due to gravity,$g = 10 \,ms^{-2}$
Since the block is pulled at a constant speed,the net force acting on the block along the inclined plane is zero.
Therefore,the tension $T$ in the rope must balance the component of the gravitational force acting down the plane:
$T = mg \sin \theta$
$T = 2 \times 10 \times \sin 30^{\circ}$
$T = 20 \times 0.5 = 10 \,N$
The work done by the tension $W$ is given by:
$W = T \times d \times \cos(0^{\circ})$
$W = 10 \,N \times 4 \,m \times 1$
$W = 40 \,J$

(B) Given the displacement equation: $s = \frac{t^2}{4}$.
Velocity of the body: $v = \frac{ds}{dt} = \frac{d}{dt}(\frac{t^2}{4}) = \frac{2t}{4} = \frac{t}{2} \ m/s$.
Acceleration of the body: $a = \frac{dv}{dt} = \frac{d}{dt}(\frac{t}{2}) = \frac{1}{2} \ m/s^2$.
Force acting on the body: $F = m \times a = 8 \ kg \times 0.5 \ m/s^2 = 4 \ N$.
At $t = 0 \ s$,displacement $s(0) = 0 \ m$.
At $t = 4 \ s$,displacement $s(4) = \frac{4^2}{4} = 4 \ m$.
Work done by the force: $W = F \times \Delta s = 4 \ N \times (4 \ m - 0 \ m) = 16 \ J$.

(C) Work done by a constant force is given by the formula:
$W = F s \cos \theta$
where $F$ is the applied force,$s$ is the displacement,and $\theta$ is the angle between the direction of force and displacement.
If $\theta = 0^{\circ}$,then $W = F s \cos 0^{\circ} = F s$ (positive).
If $\theta = 90^{\circ}$,then $W = F s \cos 90^{\circ} = 0$ (zero).
If $\theta = 180^{\circ}$,then $W = F s \cos 180^{\circ} = -F s$ (negative).
Therefore,work done can be positive,negative,or zero.

(A) The work done $W$ by a constant force $F$ is given by the dot product of the force vector and the displacement vector: $W = F \cdot s$.
Given:
$F = (5 \hat{i} + 4 \hat{j}) \text{ N}$
$s = (6 \hat{i} - 5 \hat{j} + 3 \hat{k}) \text{ m}$
Substituting these values into the formula:
$W = (5 \hat{i} + 4 \hat{j} + 0 \hat{k}) \cdot (6 \hat{i} - 5 \hat{j} + 3 \hat{k})$
$W = (5 \times 6) + (4 \times -5) + (0 \times 3)$
$W = 30 - 20 + 0$
$W = 10 \text{ J}$
Therefore,the work done by the force is $10 \text{ J}$.

(B) The displacement of the cyclist until coming to a stop is $s = 10 \,m$.
The force exerted by the road on the cycle is $F = 200 \,N$.
Since the force acts in the direction directly opposite to the motion (displacement),the angle between the force vector and the displacement vector is $\theta = 180^{\circ}$.
The work done $W$ by a constant force is given by the formula $W = F s \cos \theta$.
Substituting the given values: $W = 200 \,N \times 10 \,m \times \cos 180^{\circ}$.
Since $\cos 180^{\circ} = -1$,we get $W = 200 \times 10 \times (-1) = -2000 \,J$.
Therefore,the work done by the road on the cycle is $-2000 \,J$.