Work Done by Constant Force Questions in English

(B) The displacement vector $\Delta\vec{r}$ is given by $\Delta\vec{r} = \vec{r_2} - \vec{r_1}$.
$\Delta\vec{r} = (14\,\hat{i} + 13\,\hat{j} + 9\,\hat{k}) - (3\,\hat{i} + 2\,\hat{j} - 6\,\hat{k}) = 11\,\hat{i} + 11\,\hat{j} + 15\,\hat{k}$.
The work done $W$ is the dot product of force and displacement: $W = \vec{F} \cdot \Delta\vec{r}$.
$W = (4\,\hat{i} + \hat{j} + 3\,\hat{k}) \cdot (11\,\hat{i} + 11\,\hat{j} + 15\,\hat{k})$.
$W = (4 \times 11) + (1 \times 11) + (3 \times 15) = 44 + 11 + 45 = 100\,J$.

(D) 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 that the work done is zero,we have $\vec{F} \cdot \vec{S} = 0$.
Substituting the given vectors: $(6\hat{i} + 2\hat{j} - 3\hat{k}) \cdot (2\hat{i} - 3\hat{j} + x\hat{k}) = 0$.
Calculating the dot product: $(6 \times 2) + (2 \times -3) + (-3 \times x) = 0$.
$12 - 6 - 3x = 0$.
$6 - 3x = 0$.
$3x = 6$.
Therefore,$x = 2$.

(C) Let $T$ be the tension in the rope acting upwards and $Mg$ be the weight of the block acting downwards.
The block is moving downwards with an acceleration $a = g/2$.
Applying Newton's second law of motion: $Mg - T = Ma$
Substituting $a = g/2$,we get: $Mg - T = M(g/2)$
$T = Mg - Mg/2 = Mg/2$
The displacement $x$ is downwards,while the tension $T$ acts upwards.
Therefore,the angle between the force $T$ and the displacement $x$ is $180^{\circ}$.
Work done by the rope $W = T \cdot x \cdot \cos(180^{\circ}) = (Mg/2) \cdot x \cdot (-1) = -\frac{1}{2}Mgx$.

(D) Total mass $M = (50 + 20) \ kg = 70 \ kg$.
Total height $h = 20 \times 0.25 \ m = 5 \ m$.
The work done $W$ against gravity is given by $W = Mgh$.
Taking $g = 9.8 \ m/s^2$,we get:
$W = 70 \times 9.8 \times 5 = 3430 \ J$.

(B) The work done $W$ by a constant force $\vec{F}$ acting on a particle that undergoes a displacement $\vec{r}$ is given by the dot product of the force and displacement vectors:
$W = \vec{F} \cdot \vec{r}$
Given:
$\vec{F} = (5\hat{i} + 3\hat{j} + 2\hat{k}) \ N$
$\vec{r} = (2\hat{i} - 1\hat{j} + 0\hat{k}) \ m$
Substituting the values:
$W = (5\hat{i} + 3\hat{j} + 2\hat{k}) \cdot (2\hat{i} - 1\hat{j} + 0\hat{k})$
Using the dot product property $\hat{i} \cdot \hat{i} = 1, \hat{j} \cdot \hat{j} = 1, \hat{k} \cdot \hat{k} = 1$ and cross terms are $0$:
$W = (5 \times 2) + (3 \times -1) + (2 \times 0)$
$W = 10 - 3 + 0$
$W = 7 \ J$
Therefore,the work done is $7 \ J$.

(D) The formula for work done by a constant force is given by $W = \vec{F} \cdot \vec{S} = FS \cos \theta$.
Given values are:
Force $F = 50 \, N$
Displacement $S = 10 \, m$
Angle $\theta = 60^\circ$
Substituting these values into the formula:
$W = 50 \times 10 \times \cos 60^\circ$
Since $\cos 60^\circ = \frac{1}{2}$,we get:
$W = 500 \times \frac{1}{2} = 250 \, J$.
Therefore,the work done is $250 \, J$.

(A) The displacement vector $\vec{S}$ is given by the difference between the final position vector $\vec{r_2}$ and the initial position vector $\vec{r_1}$.
$\vec{S} = \vec{r_2} - \vec{r_1} = (14\hat i + 13\hat j + 9\hat k) - (3\hat i + 2\hat j - 6\hat k) = 11\hat i + 11\hat j + 15\hat k \, m$.
Work done $W$ is the dot product of force $\vec{F}$ and displacement $\vec{S}$.
$W = \vec{F} \cdot \vec{S} = (4\hat i + \hat j + 3\hat k) \cdot (11\hat i + 11\hat j + 15\hat k)$.
$W = (4 \times 11) + (1 \times 11) + (3 \times 15) = 44 + 11 + 45 = 100 \, J$.

(C) Work done $W$ is defined as the dot product of force $\vec{F}$ and displacement $\vec{r}$.
$W = \vec{F} \cdot \vec{r}$
Given $\vec{F} = (5\hat{i} + 3\hat{j}) \ N$ and $\vec{r} = (2\hat{i} - 1\hat{j}) \ m$.
$W = (5\hat{i} + 3\hat{j}) \cdot (2\hat{i} - 1\hat{j})$
Using the dot product property $\hat{i} \cdot \hat{i} = 1$ and $\hat{j} \cdot \hat{j} = 1$,while $\hat{i} \cdot \hat{j} = 0$:
$W = (5 \times 2) + (3 \times -1)$
$W = 10 - 3$
$W = +7 \ J$.

(A) Given: Force $F = 5 \ N$, Velocity $v = 2 \ m/s$, Time $t = 1 \ minute = 60 \ s$.
Since the velocity is constant, the displacement $d$ in time $t$ is given by $d = v \times t$.
$d = 2 \ m/s \times 60 \ s = 120 \ m$.
The work done $W$ by the force is $W = F \times d$.
$W = 5 \ N \times 120 \ m = 600 \ J$.

(D) The body starts from rest,so initial velocity $u = 0$.
Acceleration $a = \frac{v - u}{t_1} = \frac{v}{t_1}$.
Distance covered in time $t$ is given by $s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}at^2 = \frac{1}{2}at^2$.
Work done $W = F \cdot s = (ma) \cdot (\frac{1}{2}at^2) = \frac{1}{2}ma^2t^2$.
Substituting $a = \frac{v}{t_1}$,we get $W = \frac{1}{2}m \left( \frac{v}{t_1} \right)^2 t^2 = \frac{1}{2}m \frac{v^2}{t_1^2} t^2$.
Thus,the work done is proportional to $\frac{1}{2}m \frac{v^2}{t_1^2} t^2$.

(D) Let the total mass be $M$ and total length be $L$. The length of the hanging part is $l = \frac{L}{3}$.
The mass of the hanging part is $m = \frac{M}{L} \times l = \frac{M}{L} \times \frac{L}{3} = \frac{M}{3}$.
The center of mass of the hanging part is at a distance $h = \frac{l}{2} = \frac{L/3}{2} = \frac{L}{6}$ below the table surface.
The work done to pull the chain onto the table is equal to the change in potential energy of the hanging part,which is $W = mgh$.
Substituting the values: $W = (\frac{M}{3}) \times g \times (\frac{L}{6}) = \frac{MgL}{18}$.

(D) The mass of the hanging part is $m = \frac{M}{n}$,where $M = 4 \ kg$ is the total mass and $n = 4$ is the fraction of the length hanging.
So,$m = \frac{4}{4} = 1 \ kg$.
The length of the hanging part is $l = \frac{L}{n} = \frac{2}{4} = 0.5 \ m$.
The center of mass of the hanging part is at a distance $h = \frac{l}{2} = \frac{0.5}{2} = 0.25 \ m$ below the table edge.
The work done to pull the hanging part onto the table is equal to the change in potential energy of the hanging part: $W = mgh$.
Substituting the values: $W = 1 \ kg \times 10 \ m/s^2 \times 0.25 \ m = 2.5 \ J$.

(D) Given: Mass $m = 3\, kg$,displacement $s = \frac{1}{3}t^2$.
First,find the velocity $v$ by differentiating $s$ with respect to $t$:
$v = \frac{ds}{dt} = \frac{d}{dt}(\frac{1}{3}t^2) = \frac{2}{3}t$.
Next,find the acceleration $a$ by differentiating $v$ with respect to $t$:
$a = \frac{dv}{dt} = \frac{d}{dt}(\frac{2}{3}t) = \frac{2}{3}\, m/s^2$.
The force $F$ acting on the body is $F = m \times a = 3 \times \frac{2}{3} = 2\, N$.
The work done $W$ is given by $W = \int F \, ds$. Since $ds = v \, dt$,we have $W = \int_0^2 F \cdot v \, dt$.
$W = \int_0^2 2 \cdot (\frac{2}{3}t) \, dt = \frac{4}{3} \int_0^2 t \, dt$.
$W = \frac{4}{3} [\frac{t^2}{2}]_0^2 = \frac{4}{3} \times \frac{4}{2} = \frac{8}{3}\, J$.

(A) Given force $\vec{F} = (3\hat{i} + \hat{j}) \text{ N}$.
Initial position $\vec{r}_1 = (2\hat{i} + \hat{k}) \text{ m}$.
Final position $\vec{r}_2 = (4\hat{i} + 3\hat{j} - \hat{k}) \text{ m}$.
Displacement $\vec{d} = \vec{r}_2 - \vec{r}_1 = (4\hat{i} + 3\hat{j} - \hat{k}) - (2\hat{i} + \hat{k}) = (2\hat{i} + 3\hat{j} - 2\hat{k}) \text{ m}$.
Work done $W = \vec{F} \cdot \vec{d} = (3\hat{i} + \hat{j}) \cdot (2\hat{i} + 3\hat{j} - 2\hat{k})$.
$W = (3 \times 2) + (1 \times 3) + (0 \times -2) = 6 + 3 = 9 \text{ J}$.

(A) The work done $W$ by a constant force $\vec{F}$ during a displacement $\vec{d}$ is given by the dot product: $W = \vec{F} \cdot \vec{d}$.
Here,the initial position is $\vec{r}_1 = -2\hat i + 5\hat j$ and the final position is $\vec{r}_2 = 4\hat j + 3\hat k$.
The displacement vector $\vec{d} = \vec{r}_2 - \vec{r}_1 = (4\hat j + 3\hat k) - (-2\hat i + 5\hat j) = 2\hat i - \hat j + 3\hat k$.
The force is $\vec{F} = 4\hat i + 3\hat j$.
Therefore,$W = (4\hat i + 3\hat j) \cdot (2\hat i - \hat j + 3\hat k) = (4 \times 2) + (3 \times -1) + (0 \times 3) = 8 - 3 = 5 \ J$.

(A) The net force acting on the body is $\vec{F}_{net} = \vec{F}_1 + \vec{F}_2 = (2\hat i - 3\hat j + 3\hat k) + (\hat i + \hat j - 2\hat k) = 3\hat i - 2\hat j + \hat k$ $(N)$.
The displacement of the body is $\vec{d} = \vec{r}_2 - \vec{r}_1 = (7\hat i + 10\hat j + 5\hat k) - (\hat i + 2\hat j - 2\hat k) = 6\hat i + 8\hat j + 7\hat k$ $(m)$.
The work done is given by the dot product of the net force and the displacement: $W = \vec{F}_{net} \cdot \vec{d}$.
$W = (3\hat i - 2\hat j + \hat k) \cdot (6\hat i + 8\hat j + 7\hat k) = (3 \times 6) + (-2 \times 8) + (1 \times 7) = 18 - 16 + 7 = 9$ $J$.

(A) The work done $W$ by a constant force $\vec{F}$ acting on an object that undergoes a displacement $\vec{s}$ is given by the formula:
$W = \vec{F} \cdot \vec{s} = F s \cos \theta$
Given:
Force $F = 40 \,N$
Displacement $s = 8 \,m$
Angle $\theta = 60^\circ$
Substituting these values into the formula:
$W = 40 \times 8 \times \cos 60^\circ$
Since $\cos 60^\circ = 0.5$:
$W = 320 \times 0.5 = 160 \,J$
Therefore,the work done by the pulling force is $160 \,J$.

(A) Given:
Force $F = 5 \,N$
Velocity $v = 2 \,m/s$
Time $t = 1 \,minute = 60 \,s$
Since the velocity is constant,the displacement $s$ in time $t$ is given by $s = v \times t$.
$s = 2 \,m/s \times 60 \,s = 120 \,m$.
The work done $W$ by the force is $W = F \times s$.
$W = 5 \,N \times 120 \,m = 600 \,J$.

(A) The body starts from rest,so initial velocity $u = 0$.
Since the acceleration $a$ is uniform,$a = \frac{v_0 - u}{t_0} = \frac{v_0}{t_0}$.
The force acting on the body is $F = ma = m \left( \frac{v_0}{t_0} \right)$.
The distance $S$ covered in time $t$ is given by $S = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2} \left( \frac{v_0}{t_0} \right) t^2$.
The work done $W$ is equal to the product of force and displacement: $W = F \cdot S = \left( m \frac{v_0}{t_0} \right) \left( \frac{1}{2} \frac{v_0}{t_0} t^2 \right)$.
Simplifying this,we get $W = \frac{1}{2} m v_0^2 \left( \frac{t^2}{t_0^2} \right)$.

(C) The observer $B$ is on the ground (inertial frame).
For observer $B$,the block starts from rest and moves downward with an acceleration $a$.
The displacement $s$ of the block in time $t_0$ is given by the equation of motion: $s = ut + \frac{1}{2}at^2$.
Since initial velocity $u = 0$,the displacement is $s = \frac{1}{2}at_0^2$ (downward).
The force of gravity acting on the block is $F_g = mg$ (downward).
Since both the force and the displacement are in the downward direction,the work done by gravity is positive.
Work done $W = F_g \cdot s = (mg) \cdot (\frac{1}{2}at_0^2) = \frac{1}{2}mgat_0^2$.

(B) Observer $B$ is on the ground (inertial frame).
For observer $B$,the block moves downward with acceleration $a$.
The forces acting on the block are gravity ($mg$ downwards) and the normal force ($N$ upwards).
From the equation of motion: $mg - N = ma$,so $N = m(g - a)$.
The net force on the block is $F_{net} = mg - N = ma$ (downwards).
The displacement of the block in time $t_0$ is $s = \frac{1}{2}at_0^2$ (downwards).
The net work done on the block according to observer $B$ is $W = F_{net} \cdot s = (ma) \cdot (\frac{1}{2}at_0^2) = \frac{1}{2}ma^2t_0^2$.

(A) The work done by a force is given by $W = \vec{F} \cdot \vec{d} = F d cos( \theta)$,where $\theta$ is the angle between the force and the displacement vector.
In the case of a ball attached to a string,the tension force acts radially inward towards the center of rotation.
The velocity (and thus the instantaneous displacement) of the ball is always tangential to the circular path.
Since the radius is always perpendicular to the tangent,the angle between the tension force and the displacement is $90^{\circ}$.
Therefore,the work done by tension is $W = T d cos(90^{\circ}) = 0$.

(A) The work done by a constant force is given by $W = \vec{F} \cdot \vec{S}$. Since the force acts along the direction of motion,$W = F \cdot S$.
Given the total work $W_{total} = 300 \ J$ and force $F = 50 \ N$,the total distance $S_{total}$ is $S_{total} = \frac{W_{total}}{F} = \frac{300}{50} = 6 \ m$.
The motion occurs in three equal spells of distance $d$. Thus,$3d = 6 \ m$,which means $d = 2 \ m$.
The particle moves $2 \ m$ along the $x$-axis,$2 \ m$ parallel to the $y$-axis,and $2 \ m$ parallel to the $z$-axis.
Therefore,the final coordinates are $(2, 2, 2) \ m$.

(C) The force vector $\vec{F}$ is given by its magnitude multiplied by the unit vector along the direction $\hat{i} + \hat{j} + \hat{k}$.
Unit vector $\hat{n} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}$.
So,$\vec{F} = 50 \times \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}} = \frac{50}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k})\, N$.
The displacement vector $\vec{d} = \vec{r}_2 - \vec{r}_1 = (4-5)\hat{i} + (8-9)\hat{j} + (6-7)\hat{k} = -\hat{i} - \hat{j} - \hat{k}\, m$.
The work done $W = \vec{F} \cdot \vec{d} = \left[ \frac{50}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k}) \right] \cdot (-\hat{i} - \hat{j} - \hat{k})$.
$W = \frac{50}{\sqrt{3}} (-1 - 1 - 1) = \frac{50}{\sqrt{3}} (-3) = -50\sqrt{3}\, J$.

(C) The kinetic energy $K$ is given by $K = \frac{1}{2}mv^2$.
Given that kinetic energy increases uniformly with time,we have $K = kt$,where $k$ is a constant.
Thus,$\frac{1}{2}mv^2 = kt$,which implies $v^2 \propto t$,or $v \propto t^{1/2}$.
The acceleration $a$ is the rate of change of velocity: $a = \frac{dv}{dt} = \frac{d}{dt}(ct^{1/2}) = \frac{1}{2}ct^{-1/2}$.
Therefore,$a \propto \frac{1}{\sqrt{t}}$.
According to Newton's second law,the net force $F = ma$.
Since $m$ is constant,$F \propto a$,which means $F \propto \frac{1}{\sqrt{t}}$.

(A) The box is pushed at a constant velocity,so the applied force $F$ must balance the frictional force $f$.
The frictional force is given by $f = \mu N = \mu mg$.
Given: $\mu = 0.4$,$m = 40\, kg$,$g = 10\, m/s^2$.
$f = 0.4 \times 40 \times 10 = 160\, N$.
Since the box is pushed at a constant speed,the applied force $F = f = 160\, N$.
The work done by the coolies is $W = F \cdot s = 160\, N \times 20\, m = 3200\, J$.

(D) The work done $W$ by a force $F$ is given by the dot product of force and displacement: $W = \int \vec{F} \cdot d\vec{s}$.
Since the instantaneous velocity $\vec{v}$ is always tangent to the circular path and the force $\vec{F}$ (centripetal force) is always directed towards the center,the angle $\theta$ between $\vec{F}$ and the displacement $d\vec{s}$ (which is in the direction of $\vec{v}$) is always $90^{\circ}$.
Therefore,the work done for an infinitesimal displacement is $dW = F \cdot ds \cdot \cos(90^{\circ}) = 0$.
Integrating this over one complete rotation,the total work done is $W = 0$.

(C) The mass per unit length of the chain is $\lambda = \frac{M}{L} = \frac{4\,kg}{2\,m} = 2\,kg/m$.
The mass of the hanging part is $m = \lambda \times \ell = 2\,kg/m \times 0.6\,m = 1.2\,kg$.
The center of mass of the hanging part is at a distance of $h = \frac{\ell}{2} = \frac{0.6\,m}{2} = 0.3\,m$ below the edge of the table.
The work done $W$ to pull the hanging part onto the table is equal to the change in potential energy of the hanging part,which is $W = mgh = (1.2\,kg) \times (10\,m/s^2) \times (0.3\,m) = 3.6\,J$.

(C) The displacement of the box with respect to the ground is the vector sum of its displacement relative to the train and the displacement of the train relative to the ground.
Therefore,the total displacement $\vec{S}_{total} = \vec{s} + \vec{s}_0$.
Work done is defined as the dot product of the force applied and the total displacement in the reference frame.
$W = \vec{F} \cdot \vec{S}_{total} = \vec{F} \cdot (\vec{s} + \vec{s}_0)$.

(A) Given,kinetic energy $K = as^2$.
We know that $K = \frac{1}{2}mv^2$,where $m$ is the mass and $v$ is the velocity of the particle.
So,$\frac{1}{2}mv^2 = as^2$.
Differentiating both sides with respect to time $t$:
$\frac{d}{dt}(\frac{1}{2}mv^2) = \frac{d}{dt}(as^2)$
$\frac{1}{2}m(2v \frac{dv}{dt}) = a(2s \frac{ds}{dt})$
Since $F = m \frac{dv}{dt}$ and $v = \frac{ds}{dt}$,we substitute these into the equation:
$mv \frac{dv}{dt} = 2as \frac{ds}{dt}$
$F \cdot v = 2as \cdot v$
$F = 2as$.

(A) The given force $\vec{F} = (3\hat{i} + 4\hat{j})$ is a constant force.
$A$ constant force is always a conservative force.
For a conservative force,the work done is independent of the path taken and depends only on the initial and final positions of the particle.
Since both paths start at $(0, 0)$ and end at $(a, a)$,the work done in both cases must be equal.
Therefore,$W_1 = W_2$.

(D) Given,initial velocity $u = 0$,final velocity $v$ at time $t_1$.
Acceleration $a = \frac{v - u}{t_1} = \frac{v}{t_1}$.
Work done $W$ at any time $t$ is equal to the change in kinetic energy or $F \times s$.
Using $s = \frac{1}{2} a t^2$ and $F = ma$:
$W = F \times s = (ma) \times (\frac{1}{2} a t^2) = \frac{1}{2} m a^2 t^2$.
Substituting $a = \frac{v}{t_1}$:
$W = \frac{1}{2} m \left(\frac{v}{t_1}\right)^2 t^2 = \frac{1}{2} m \frac{v^2}{t_1^2} t^2$.

(D) Work done is defined as the dot product of force and displacement,given by $W = \vec{F} \cdot \vec{s} = Fs \cos \theta$,where $\theta$ is the angle between the force and displacement vectors.
In uniform circular motion,the centripetal force acts towards the center of the circle,while the instantaneous displacement is along the tangent to the circular path.
Since the radius is always perpendicular to the tangent,the angle between the centripetal force and the displacement is $90^{\circ}$.
Therefore,$W = Fs \cos(90^{\circ}) = Fs(0) = 0$.
Thus,no work is done by the centripetal force.

(D) The acceleration of the body is $a = \frac{v}{t_1}$.
The force acting on the body is $F = m \cdot a = m \cdot \frac{v}{t_1}$.
The distance covered by the body in time $t$ starting from rest is $s = \frac{1}{2} a t^2 = \frac{1}{2} \left( \frac{v}{t_1} \right) t^2$.
Work done is defined as $W = F \cdot s$.
Substituting the values,$W = \left( m \cdot \frac{v}{t_1} \right) \cdot \left( \frac{1}{2} \cdot \frac{v}{t_1} \cdot t^2 \right)$.
Simplifying the expression,$W = \frac{1}{2} m \left( \frac{v}{t_1} \right)^2 t^2 = \frac{1}{2} m \frac{v^2}{t_1^2} t^2$.

(B) Let the total length of the chain be $L = 2 \, m$ and its total mass be $M = 4 \, kg$.
The length of the chain hanging from the table is $l = 60 \, cm = 0.6 \, m$.
The mass per unit length of the chain is $\lambda = \frac{M}{L} = \frac{4}{2} = 2 \, kg/m$.
The mass of the hanging part is $m = \lambda \times l = 2 \times 0.6 = 1.2 \, kg$.
The center of mass of the hanging part is at a distance $h = \frac{l}{2} = \frac{0.6}{2} = 0.3 \, m$ below the table edge.
The work done to pull the chain onto the table is equal to the change in potential energy of the hanging part,which is $W = mgh$.
Substituting the values: $W = 1.2 \times 10 \times 0.3 = 3.6 \, J$.

(A) According to the work-energy theorem,the change in kinetic energy of a particle is equal to the work done by the net force acting on it.
$\Delta K = W = \vec{F} \cdot \Delta \vec{r}$
Given:
$\vec{F} = (7\hat{i} + 4\hat{j} + 3\hat{k}) \text{ N}$
$\Delta \vec{r} = (2\hat{i} + 3\hat{j} + 4\hat{k}) \text{ m}$
Calculating the dot product:
$W = (7 \times 2) + (4 \times 3) + (3 \times 4)$
$W = 14 + 12 + 12$
$W = 38 \text{ J}$
Therefore,the change in kinetic energy is $38 \text{ J}$.

(B) The total work done is the sum of the work done to lift the bucket,the water,and the rope.
Mass of bucket $(m_b)$ = $2\,kg$.
Mass of water $(m_w)$ = $8\,litre \times 1\,kg/litre = 8\,kg$.
Mass of rope $(m_r)$ = $1\,kg$.
Depth $(h)$ = $5\,m$.
Work done on bucket: $W_b = m_b \cdot g \cdot h = 2 \times 10 \times 5 = 100\,J$.
Work done on water: $W_w = m_w \cdot g \cdot h = 8 \times 10 \times 5 = 400\,J$.
Work done on rope: Since the rope is uniform,its center of mass is at $h/2 = 2.5\,m$. Thus,$W_r = m_r \cdot g \cdot (h/2) = 1 \times 10 \times 2.5 = 25\,J$.
Total work $W = W_b + W_w + W_r = 100 + 400 + 25 = 525\,J$.

(C) The force acting on the particle is given by $F = \frac{K}{v}$.
We know that the work done $W$ is given by the integral of power over time,or $W = \int F \cdot dx$.
Since $v = \frac{dx}{dt}$,we can write $dx = v \cdot dt$.
Substituting this into the work equation:
$W = \int F \cdot (v \cdot dt)$
Substitute $F = \frac{K}{v}$ into the equation:
$W = \int \left(\frac{K}{v}\right) \cdot v \cdot dt$
$W = \int K \cdot dt$
Since $K$ is a constant,the work done in time $t$ is:
$W = K \int_{0}^{t} dt = Kt$.

(A) 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 vector.
Given:
Force $F = 40\, N$
Displacement $S = 8\, m$
Angle $\theta = 60^{\circ}$
Substituting these values into the formula:
$W = 40 \times 8 \times \cos 60^{\circ}$
Since $\cos 60^{\circ} = 0.5$ or $\frac{1}{2}$,we have:
$W = 40 \times 8 \times 0.5$
$W = 320 \times 0.5 = 160\, J$
Therefore,the work done is $160\, J$.

(B) The force acting on the body is given by $\vec{F} = -\hat{i} + 2\hat{j} + 3\hat{k} \text{ N}$.
Since the body is constrained to move along the $y$-axis for a distance of $4 \text{ m}$,the displacement vector is $\vec{d} = 4\hat{j} \text{ m}$.
The work done $W$ by a constant force is defined as the dot product of the force vector and the displacement vector: $W = \vec{F} \cdot \vec{d}$.
Substituting the values: $W = (-\hat{i} + 2\hat{j} + 3\hat{k}) \cdot (4\hat{j})$.
Using the dot product properties ($\hat{i} \cdot \hat{j} = 0$,$\hat{j} \cdot \hat{j} = 1$,$\hat{k} \cdot \hat{j} = 0$):
$W = (-1 \times 0) + (2 \times 4) + (3 \times 0) = 8 \text{ J}$.
Thus,the work done is $8 \text{ J}$.

(C) The force of gravity acting on the particle is $\vec{F} = -mg \hat{j}$.
At the highest point,the vertical displacement of the particle is equal to the maximum height $H$.
The formula for maximum height is $H = \frac{u^2 \sin^2 \alpha}{2g}$.
The work done by gravity is given by the product of the force and the vertical displacement: $W = \vec{F} \cdot \vec{S} = (-mg) \times H$.
Substituting the value of $H$: $W = -mg \left( \frac{u^2 \sin^2 \alpha}{2g} \right)$.
Simplifying the expression,we get $W = -\frac{m u^2 \sin^2 \alpha}{2}$.

(B) The force acting on the body is the centripetal force,which is always directed towards the center of the circular path.
The displacement of the body at any instant is along the tangent to the circle at that point.
Since the radius (and thus the centripetal force) is always perpendicular to the tangent (displacement),the angle $\theta$ between the force vector $\vec{F}$ and the displacement vector $d\vec{s}$ is always $90^{\circ}$.
The work done $W$ is given by the dot product of force and displacement: $W = \int \vec{F} \cdot d\vec{s} = \int F \cos(90^{\circ}) ds$.
Since $\cos(90^{\circ}) = 0$,the work done by the centripetal force is always zero,regardless of the distance traveled along the circumference.

(B) The work done $W$ by a constant force $\vec F$ during a displacement $\vec S$ is given by the dot product of the force and displacement vectors:
$W = \vec F \cdot \vec S$
Given $\vec F = 2\hat i - 3\hat j + 7\hat k$ and $\vec S = 7\hat i + 3\hat j - 2\hat k$.
$W = (2\hat i - 3\hat j + 7\hat k) \cdot (7\hat i + 3\hat j - 2\hat k)$
$W = (2 \times 7) + (-3 \times 3) + (7 \times -2)$
$W = 14 - 9 - 14$
$W = -9 \text{ J}$

(A) The work done by a force is given by the formula $W = F \cdot d \cdot \cos(\theta)$,where $F$ is the force,$d$ is the displacement,and $\theta$ is the angle between the force vector and the displacement vector.
In this case,the weight (gravitational force) acts vertically downwards,i.e.,$F = Mg$ (downwards).
The displacement $x$ is along the horizontal table,i.e.,in the horizontal direction.
The angle $\theta$ between the downward gravitational force and the horizontal displacement is $90^{\circ}$.
Therefore,the work done by the weight is $W = Mg \cdot x \cdot \cos(90^{\circ})$.
Since $\cos(90^{\circ}) = 0$,the work done $W = 0$.

(C) The work done $W$ is defined as the dot product of the force vector $\vec F$ and the displacement vector $\vec s$,given by $W = \vec F \cdot \vec s = Fs \cos \theta$,where $\theta$ is the angle between the two vectors.
Given that $W = 0$,$F \neq 0$,and $s \neq 0$,we have $Fs \cos \theta = 0$.
Since $F$ and $s$ are non-zero,it must be that $\cos \theta = 0$.
This implies $\theta = 90^\circ$ or $\pi/2$ radians.
Therefore,the force vector $\vec F$ is perpendicular to the displacement vector $\vec s$.

(C) According to the Work-Energy Theorem,the work done by a force on a body is equal to the change in its kinetic energy: $W = \Delta KE = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2$.
For the velocity to increase,the final kinetic energy must be greater than the initial kinetic energy,which implies that the work done $W$ must be positive.
The work done is given by the dot product $W = \vec{F} \cdot \vec{d} = F d \cos \theta$,where $\theta$ is the angle between the force vector $\vec{F}$ and the displacement vector $\vec{d}$.
If $\hat{F} \parallel \hat{d}$,then the angle $\theta = 0^\circ$,and $\cos 0^\circ = 1$.
Thus,$W = F d > 0$,which results in an increase in kinetic energy and consequently an increase in the velocity of the body.
Therefore,the correct condition is $\hat{F} \parallel \hat{d}$.

(C) In physics,work is defined as the product of force and displacement in the direction of the force $(W = Fs \cos \theta)$.
$(i)$ When the weightlifter lifts the weight,there is a displacement $(s > 0)$ in the direction of the applied force. Therefore,work is done $(W = Fs > 0)$.
$(ii)$ When the weightlifter holds the weight stationary,the displacement is zero $(s = 0)$. Since work requires displacement,the work done while holding the weight is zero $(W = F \times 0 = 0)$.

(A) The work done by a constant force acting in the direction of displacement is given by $W = F \cdot d$,where $d$ is the distance covered.
Since the force $F = 50 \, N$ acts along the direction of motion in each segment,the work done in each segment is $W_i = F \cdot d$.
Given that the total work done is $W_{total} = 300 \, J$ and the work done in each of the three segments is equal,the work done in each segment is $W_i = \frac{300}{3} = 100 \, J$.
Using $W_i = F \cdot d$,we find the distance $d$ for each segment: $100 = 50 \cdot d$,which gives $d = 2 \, m$.
The particle moves $2 \, m$ along the $x$-axis,then $2 \, m$ parallel to the $y$-axis,and finally $2 \, m$ parallel to the $z$-axis.
Starting from the origin $(0, 0, 0)$,the final coordinates are $(2, 2, 2) \, m$.

(B) The block $m$ moves with the same horizontal acceleration $a = 5\,m/s^2$ as the wedge because it does not slip on the wedge.
Since the surfaces are frictionless,the only force acting on the block in the horizontal direction is the horizontal component of the normal force $N$.
Let $N$ be the normal force exerted by the wedge on the block. The horizontal component of $N$ provides the acceleration $a$ to the block: $N \sin \theta = ma$.
The vertical component of $N$ balances the weight: $N \cos \theta = mg$.
However,we can use the Work-Energy Theorem directly. The work done by all forces on the block equals the change in its kinetic energy.
The only force doing work on the block in the horizontal direction is the normal force $N$. Since the block moves only horizontally with acceleration $a$,its velocity at time $t = 2\,s$ is $v = at = 5 \times 2 = 10\,m/s$.
The initial velocity is $0$. The change in kinetic energy is $\Delta KE = \frac{1}{2}mv^2 - 0 = \frac{1}{2} \times 1 \times (10)^2 = 50\,J$.
Since the normal force is the only force doing work in the direction of motion (as gravity acts vertically and displacement is horizontal),the work done by the normal force is $50\,J$.

(C) The force vector $\vec{F}$ is given by its magnitude multiplied by the unit vector in the direction of $\hat{i} + \hat{j} + \hat{k}$.
Unit vector $\hat{n} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}$.
So,$\vec{F} = 30 \times \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}} = 10\sqrt{3}(\hat{i} + \hat{j} + \hat{k})\, N$.
The displacement vector $\vec{S}$ is the difference between the final and initial position vectors:
$\vec{S} = (3-2)\hat{i} + (5-4)\hat{j} + (2-1)\hat{k} = \hat{i} + \hat{j} + \hat{k}\, m$.
The work done $W$ is the dot product of $\vec{F}$ and $\vec{S}$:
$W = \vec{F} \cdot \vec{S} = [10\sqrt{3}(\hat{i} + \hat{j} + \hat{k})] \cdot [\hat{i} + \hat{j} + \hat{k}]$
$W = 10\sqrt{3} (1 \times 1 + 1 \times 1 + 1 \times 1) = 10\sqrt{3} \times 3 = 30\sqrt{3}\, J$.