Work Done by Constant Force Questions in English

(D) The work done $W$ by a constant force $\overrightarrow{F}$ causing a displacement $\overrightarrow{S}$ is given by the dot product: $W = \overrightarrow{F} \cdot \overrightarrow{S} = FS \cos \theta$.
Given: Force $F = 50 \, N$,displacement $S = 10 \, m$,and angle $\theta = 60^\circ$.
Substituting these values into the formula: $W = 50 \times 10 \times \cos 60^\circ$.
Since $\cos 60^\circ = 0.5$,we have $W = 500 \times 0.5 = 250 \, J$.

(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) The work done $W$ by a constant force $\vec F$ during a displacement $\vec r$ is given by the dot product of the force and displacement vectors:
$W = \vec F \cdot \vec r$
Given $\vec F = (5\hat i + 3\hat j) \text{ N}$ and $\vec r = (2\hat i - 1\hat j) \text{ m}$.
Substituting the values:
$W = (5\hat i + 3\hat j) \cdot (2\hat i - 1\hat j)$
Using the properties of unit vectors $(\hat i \cdot \hat i = 1, \hat j \cdot \hat j = 1, \hat i \cdot \hat j = 0)$:
$W = (5 \times 2) + (3 \times -1)$
$W = 10 - 3$
$W = 7 \text{ J}$.
Thus,the correct option is $C$.

(A) The work done $W$ by a constant force $\overrightarrow{F}$ is given by the dot product of the force vector and the displacement vector $\overrightarrow{S}$.
$W = \overrightarrow{F} \cdot \overrightarrow{S}$
Given $\overrightarrow{F} = 5\hat{i} + 6\hat{j} + 4\hat{k}$ and $\overrightarrow{S} = 6\hat{i} + 0\hat{j} - 5\hat{k}$.
$W = (5\hat{i} + 6\hat{j} + 4\hat{k}) \cdot (6\hat{i} + 0\hat{j} - 5\hat{k})$
Using the property $\hat{i} \cdot \hat{i} = 1$,$\hat{j} \cdot \hat{j} = 1$,$\hat{k} \cdot \hat{k} = 1$ and cross-terms are $0$:
$W = (5 \times 6) + (6 \times 0) + (4 \times -5)$
$W = 30 + 0 - 20$
$W = 10 \ J$.

(B) The work done $W$ by a constant force $\vec{F}$ during a displacement $\vec{r}$ is given by the dot product: $W = \vec{F} \cdot \vec{r}$.
Given the force $\vec{F} = (-2\hat{i} + 15\hat{j} + 6\hat{k})\,N$.
The displacement is $10\,m$ along the $Y$-axis,so $\vec{r} = 10\hat{j}\,m$.
Substituting these into the formula:
$W = (-2\hat{i} + 15\hat{j} + 6\hat{k}) \cdot (10\hat{j})$
$W = (-2 \times 0) + (15 \times 10) + (6 \times 0)$
$W = 0 + 150 + 0 = 150\,J$.
Therefore,the work done is $150\,J$.

(A) Work done $W$ is defined as the dot product of force $\overrightarrow F$ and displacement $\overrightarrow s$.
$W = \overrightarrow F \cdot \overrightarrow s$
Given $\overrightarrow F = 4\hat i + 5\hat j$ and $\overrightarrow s = 3\hat i + 6\hat k$.
$W = (4\hat i + 5\hat j) \cdot (3\hat i + 6\hat k)$
Using the properties of unit vectors $\hat i \cdot \hat i = 1$, $\hat j \cdot \hat j = 1$, $\hat k \cdot \hat k = 1$, and $\hat i \cdot \hat j = \hat j \cdot \hat k = \hat k \cdot \hat i = 0$:
$W = (4 \times 3)(\hat i \cdot \hat i) + (4 \times 6)(\hat i \cdot \hat k) + (5 \times 3)(\hat j \cdot \hat i) + (5 \times 6)(\hat j \cdot \hat k)$
$W = 12(1) + 0 + 0 + 0 = 12 \text{ J}$.

(A) 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 $\vec F = 3\hat i + c\hat j + 2\hat k$ and $\vec S = -4\hat i + 2\hat j - 3\hat k$.
Substituting these into the formula: $W = (3\hat i + c\hat j + 2\hat k) \cdot (-4\hat i + 2\hat j - 3\hat k)$.
Calculating the dot product: $W = (3)(-4) + (c)(2) + (2)(-3) = -12 + 2c - 6 = 2c - 18$.
Given that the work done $W = 6 \ J$,we set up the equation: $2c - 18 = 6$.
Adding $18$ to both sides: $2c = 24$.
Dividing by $2$: $c = 12$.

(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 $\vec F = (5\hat i + 3\hat j) \ N$ and $\vec s = (2\hat i - 1\hat j) \ m$.
$W = (5\hat i + 3\hat j) \cdot (2\hat i - 1\hat j)$
Using the dot product properties $\hat i \cdot \hat i = 1$,$\hat j \cdot \hat j = 1$,and $\hat i \cdot \hat j = 0$:
$W = (5 \times 2) + (3 \times -1)$
$W = 10 - 3$
$W = 7 \ J$
Therefore,the correct option is $B$.

(A) In a horizontal circular motion,the centripetal force acts towards the center of the circle,while the displacement of the sphere at any instant is along the tangent to the circular path.
Since the centripetal force is always perpendicular to the direction of displacement (velocity vector),the angle $\theta$ between the force and the displacement is $90^{\circ}$.
The work done $W$ is given by the formula $W = F \cdot s \cdot \cos(\theta)$.
Since $\cos(90^{\circ}) = 0$,the work done by the centripetal force in one full horizontal circle is $0$.

(D) The work done by a force is given by the formula $W = \vec{F} \cdot \vec{d} = F d \cos \theta$,where $\theta$ is the angle between the force and the displacement.
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 vector is always perpendicular to the tangent,the angle between the centripetal force and the displacement is always $90^\circ$.
Therefore,the work done in any small displacement is $dW = F \cdot ds \cdot \cos(90^\circ) = 0$.
Consequently,the total work done in one complete revolution is $0 \, J$.

(C) Given: Mass $m = 3 \times 10^7 \, kg$,Force $F = 5 \times 10^4 \, N$,Distance $s = 3 \, m$,Initial velocity $u = 0$.
According to Newton's second law,acceleration $a = \frac{F}{m} = \frac{5 \times 10^4}{3 \times 10^7} = \frac{5}{3} \times 10^{-3} \, m/s^2$.
Using the kinematic equation $v^2 - u^2 = 2as$:
$v^2 = 0^2 + 2 \times \left( \frac{5 \times 10^4}{3 \times 10^7} \right) \times 3$.
$v^2 = 2 \times \left( \frac{5}{3} \times 10^{-3} \right) \times 3 = 10 \times 10^{-3} = 10^{-2}$.
$v = \sqrt{10^{-2}} = 0.1 \, m/s$.

(C) The formula for work done is given by $W = Fs \cos \theta$,where $W$ is work,$F$ is force,$s$ is displacement,and $\theta$ is the angle between the force and the direction of motion.
Given: $W = 25 \, J$,$F = 5 \, N$,and $s = 10 \, m$.
Substituting the values into the formula:
$25 = 5 \times 10 \times \cos \theta$
$25 = 50 \times \cos \theta$
$\cos \theta = \frac{25}{50} = \frac{1}{2}$
Since $\cos 60^\circ = \frac{1}{2}$,the angle $\theta = 60^\circ$.

(B) The work done $(W)$ by an external force in lifting an object against gravity is given by the formula: $W = mgh$.
Here,$m$ is the mass of the object,$g$ is the acceleration due to gravity,and $h$ is the vertical displacement (height).
Since the weight of the book is $W_b = mg$,the work done can be expressed as $W = W_b \times h$.
Therefore,the work done depends only on the weight of the book and the height of the bookshelf.
It does not depend on the time taken to perform the work,as time is a factor in calculating power,not work.

(B) The work done $W$ in lifting a body of mass $m$ to a height $h$ is given by the formula $W = mgh$.
Here,the mass $m$ and the height $h$ are the same for both men.
Since work done is independent of the time taken to perform the task,the work done by both men is equal.
Therefore,the ratio of the work done is $W_1 : W_2 = mgh : mgh = 1:1$.

(C) The work done $W$ by a constant force $\overrightarrow F$ during a displacement $\overrightarrow s$ is given by the dot product: $W = \overrightarrow F \cdot \overrightarrow s$.
Given: $\overrightarrow F = (5\hat i + 3\hat j) \, N$ and $\overrightarrow s = (2\hat i - 1\hat j) \, m$.
Substituting the values: $W = (5\hat i + 3\hat j) \cdot (2\hat i - 1\hat j)$.
Using the dot product property $\hat i \cdot \hat i = 1$ and $\hat i \cdot \hat j = 0$:
$W = (5 \times 2) + (3 \times -1) = 10 - 3 = 7 \, J$.

(D) The work done by a constant force is given by the formula $W = F \cdot s \cdot \cos(\theta)$.
Here,the gravitational force $F = mg$ acts downwards,and the displacement $s = h$ is also in the downward direction.
Since the force and displacement are in the same direction,the angle $\theta = 0^\circ$,and $\cos(0^\circ) = 1$.
Therefore,the work done is $W = mgh$.
Given $m = 10\, kg$,$g = 9.8\, m/s^2$,and $h = 10\, m$.
$W = 10 \times 9.8 \times 10 = 980\, J$.
Since the body moves in the direction of the gravitational force,the work done is positive.

(D) The correct answer is $(d)$.
Work is defined as the dot product of force and displacement, represented as $W = \vec{F} \cdot \vec{s}$.
Since the dot product of two vectors always results in a scalar, work is a scalar quantity.
Displacement, electric field, and acceleration are all vector quantities because they possess both magnitude and direction.

(B) The force required to pull the block up the smooth inclined plane at a constant velocity is equal to the component of the gravitational force acting down the plane,which is $F = mg \sin \theta$.
Given that the weight of the block $mg = 2 \, kN = 2000 \, N$,the length of the plane $s = 10 \, m$,and the angle of inclination $\theta = 15^\circ$.
The work done $W$ is given by the product of the force and the displacement: $W = F \times s = (mg \sin \theta) \times s$.
Substituting the values: $W = 2000 \times \sin(15^\circ) \times 10$.
Since $\sin(15^\circ) \approx 0.2588$,we get $W = 2000 \times 0.2588 \times 10 = 5176 \, J$.
Converting to kilojoules: $W = 5.176 \, kJ \approx 5.17 \, kJ$.

(D) 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 $\vec F = 5\hat i + 6\hat j - 4\hat k$ and $\vec s = 6\hat i + 0\hat j + 5\hat k$.
$W = (5\hat i + 6\hat j - 4\hat k) \cdot (6\hat i + 0\hat j + 5\hat k)$
$W = (5 \times 6) + (6 \times 0) + (-4 \times 5)$
$W = 30 + 0 - 20$
$W = 10 \text{ units}$.

(B) Given: Force $F = 5 \, N$,mass $m = 15 \, kg$,initial velocity $u = 0$,time $t = 1 \, s$.
First,calculate the acceleration $a$ using Newton's second law: $a = \frac{F}{m} = \frac{5}{15} = \frac{1}{3} \, m/s^2$.
Next,calculate the displacement $s$ in the first second using the equation of motion: $s = ut + \frac{1}{2}at^2 = 0(1) + \frac{1}{2} \times \frac{1}{3} \times (1)^2 = \frac{1}{6} \, m$.
Finally,calculate the work done $W$ using the formula $W = F \times s$: $W = 5 \times \frac{1}{6} = \frac{5}{6} \, J$.

(B) The work done against gravity is given by the formula $W = mgh$.
Here,mass $m = 10 \, kg$,acceleration due to gravity $g = 9.8 \, m/s^2$,and height $h = 1 \, m$.
Substituting the values,we get $W = 10 \times 9.8 \times 1 = 98 \, J$.
The time taken is irrelevant for calculating the work done against gravity.
Therefore,the correct option is $B$.

(D) Given displacement $S = \frac{t^2}{4}$.
Velocity $v = \frac{dS}{dt} = \frac{d}{dt}\left(\frac{t^2}{4}\right) = \frac{2t}{4} = \frac{t}{2} \, m/s$.
Acceleration $a = \frac{dv}{dt} = \frac{d}{dt}\left(\frac{t}{2}\right) = \frac{1}{2} \, m/s^2$.
Force $F = m \cdot a = 6 \times \frac{1}{2} = 3 \, N$.
Work done $W = \int F \, dS = \int_0^2 F \cdot v \, dt = \int_0^2 3 \cdot \frac{t}{2} \, dt$.
$W = \frac{3}{2} \int_0^2 t \, dt = \frac{3}{2} \left[ \frac{t^2}{2} \right]_0^2 = \frac{3}{4} [4 - 0] = 3 \, J$.

(D) The work done $W$ by a constant force $\overrightarrow{F}$ during a displacement $\overrightarrow{s}$ is given by the dot product of the force and displacement vectors:
$W = \overrightarrow{F} \cdot \overrightarrow{s}$
Given:
$\overrightarrow{F} = (3\,\hat{i} + 4\,\hat{j}) \, N$
$\overrightarrow{s} = (3\,\hat{i} + 4\,\hat{j}) \, m$
Substituting the values:
$W = (3\,\hat{i} + 4\,\hat{j}) \cdot (3\,\hat{i} + 4\,\hat{j})$
Using the dot product property $\hat{i} \cdot \hat{i} = 1$,$\hat{j} \cdot \hat{j} = 1$,and $\hat{i} \cdot \hat{j} = 0$:
$W = (3 \times 3) + (4 \times 4)$
$W = 9 + 16$
$W = 25 \, J$

(D) The total mass $m$ being lifted is the sum of the man's mass and the load: $m = 50\,kg + 20\,kg = 70\,kg$.
The total height $h$ climbed is the number of steps multiplied by the height of each step: $h = 20 \times 0.25\,m = 5\,m$.
The work done $W$ against gravity is given by the formula $W = mgh$,where $g$ is the acceleration due to gravity $(g \approx 9.8\,m/s^2)$.
Substituting the values: $W = 70\,kg \times 9.8\,m/s^2 \times 5\,m = 3430\,J$.

(D) The work done $W$ by a force $\overrightarrow{F}$ during a displacement $\overrightarrow{s}$ is given by the dot product: $W = \overrightarrow{F} \cdot \overrightarrow{s}$.
Given $\overrightarrow{F} = 6\hat{i} + 2\hat{j} - 3\hat{k}$ and $\overrightarrow{s} = 2\hat{i} - 3\hat{j} + x\hat{k}$.
Since the work done is zero,we have $W = 0$.
Substituting the vectors into the dot product formula:
$(6\hat{i} + 2\hat{j} - 3\hat{k}) \cdot (2\hat{i} - 3\hat{j} + x\hat{k}) = 0$
Calculating the dot product:
$(6 \times 2) + (2 \times -3) + (-3 \times x) = 0$
$12 - 6 - 3x = 0$
$6 - 3x = 0$
$3x = 6$
$x = 2$.

(A) The displacement $\vec{d}$ is given by $\vec{d} = \vec{r}_2 - \vec{r}_1$.
$\vec{d} = (14\hat{i} + 13\hat{j} + 9\hat{k}) - (3\hat{i} + 2\hat{j} - 6\hat{k}) = 11\hat{i} + 11\hat{j} + 15\hat{k}$.
The work done $W$ is the dot product of force and displacement: $W = \vec{F} \cdot \vec{d}$.
$W = (4\hat{i} + \hat{j} + 3\hat{k}) \cdot (11\hat{i} + 11\hat{j} + 15\hat{k})$.
$W = (4 \times 11) + (1 \times 11) + (3 \times 15) = 44 + 11 + 45 = 100 \text{ J}$.

(C) The work done $W$ by a constant force $\vec{F}$ during a displacement $\vec{s}$ is given by the dot product: $W = \vec{F} \cdot \vec{s}$.
Given $\vec{F} = 3\hat{i} + c\hat{j} + 2\hat{k}$ and $\vec{s} = -4\hat{i} + 2\hat{j} + 3\hat{k}$,and $W = 6\,J$.
Substituting the values into the dot product formula:
$6 = (3\hat{i} + c\hat{j} + 2\hat{k}) \cdot (-4\hat{i} + 2\hat{j} + 3\hat{k})$
$6 = (3 \times -4) + (c \times 2) + (2 \times 3)$
$6 = -12 + 2c + 6$
$6 = -6 + 2c$
$12 = 2c$
$c = 6$.

(B) The formula for work done by a constant force is $W = F \cdot d$,where $F$ is the force and $d$ is the displacement in the direction of the force.
If the force is doubled $(F' = 2F)$ and the displacement is doubled $(d' = 2d)$,the new work done $W'$ is:
$W' = F' \cdot d' = (2F) \cdot (2d) = 4 \cdot (F \cdot d) = 4W$.
Therefore,the work done becomes $4$ times the original work.

(D) The work done by a constant force is given by the formula $W = F S \cos \theta$,where $F$ is the magnitude of the force,$S$ is the displacement,and $\theta$ is the angle between the force vector and the displacement vector.
Given:
Force $F = 10\, N$
Displacement $S = 4\, m$
Angle $\theta = 60^\circ$
Substituting these values into the formula:
$W = 10 \times 4 \times \cos 60^\circ$
Since $\cos 60^\circ = 0.5$,we have:
$W = 10 \times 4 \times 0.5 = 20\, J$.
Therefore,the work done by the force is $20\, J$.

(A) The work done $W$ is given by the dot product of the force vector $\overrightarrow F$ and the displacement vector $\overrightarrow S$.
$W = \overrightarrow F \cdot \overrightarrow S$
Given $\overrightarrow F = (5\hat i + 4\hat j + 0\hat k) \ N$ and $\overrightarrow S = (6\hat i - 5\hat j + 3\hat k) \ m$.
$W = (5\hat i + 4\hat j + 0\hat k) \cdot (6\hat i - 5\hat j + 3\hat k)$
$W = (5 \times 6) + (4 \times -5) + (0 \times 3)$
$W = 30 - 20 + 0$
$W = 10 \ J$.

(B) The work done $W$ by a constant force $\vec F$ during a displacement $\vec r$ is given by the dot product of the force and displacement vectors:
$W = \vec F \cdot \vec r$
Given $\vec F = (5\hat i + 3\hat j + 2\hat k) \, N$ and $\vec r = (2\hat i - \hat j + 0\hat k) \, m$.
$W = (5\hat i + 3\hat j + 2\hat k) \cdot (2\hat i - 1\hat j + 0\hat k)$
$W = (5 \times 2) + (3 \times -1) + (2 \times 0)$
$W = 10 - 3 + 0 = 7 \, J$
Therefore,the work done is $+7 \, J$.

(A) According to the work-energy theorem,the kinetic energy $(K.E.)$ acquired by a body is equal to the work done on it by the constant force $F$.
$K.E. = W = F \times s$
Since the force $F$ and the distance $s$ are given as constants,the work done $W$ is constant.
Therefore,the kinetic energy $K.E.$ is independent of the mass $m$ of the body.
Mathematically,$K.E. \propto m^0$.
Thus,the correct option is $A$.

(D) The work done $W$ is given by the dot product of the force vector $\vec F$ and the displacement vector $\vec s$.
$W = \vec F \cdot \vec s$
Given $\vec F = 4\hat i + 5\hat j + 0\hat k$ and $\vec s = 3\hat i + 0\hat j + 6\hat k$.
$W = (4\hat i + 5\hat j + 0\hat k) \cdot (3\hat i + 0\hat j + 6\hat k)$
Using the property of dot products $\hat i \cdot \hat i = 1$,$\hat j \cdot \hat j = 1$,$\hat k \cdot \hat k = 1$,and cross terms are $0$:
$W = (4 \times 3) + (5 \times 0) + (0 \times 6)$
$W = 12 + 0 + 0 = 12$ units.
Comparing this with the options,$4 \times 3 = 12$ units.
Therefore,the correct option is $D$.

(A) The work done by the man is $W = \int \vec{F} \cdot d\vec{r}$.
Since the surface is assumed to be smooth,there is no friction force acting on the man.
The only force acting on the man is the gravitational force,which always acts towards the center of the Earth.
As the man moves along the surface of the Earth,his displacement vector $d\vec{r}$ is always tangent to the surface,while the gravitational force $\vec{F}_g$ is always directed towards the center (normal to the surface).
Since the angle between the displacement vector and the gravitational force is $90^{\circ}$ at every point,the work done by gravity is $W = \int F_g \cos(90^{\circ}) dr = 0$.
Therefore,the net work done by the man is $0$.

(A) Work done $W$ is defined as the dot product of force vector $\overrightarrow{F}$ and displacement vector $\overrightarrow{s}$.
$W = \overrightarrow{F} \cdot \overrightarrow{s}$
Given $\overrightarrow{F} = (6\hat{i} + 2\hat{j}) \text{ N}$ and $\overrightarrow{s} = (3\hat{i} - \hat{j}) \text{ m}$.
$W = (6\hat{i} + 2\hat{j}) \cdot (3\hat{i} - \hat{j})$
Using the dot product property $\hat{i} \cdot \hat{i} = 1$,$\hat{j} \cdot \hat{j} = 1$,and $\hat{i} \cdot \hat{j} = 0$:
$W = (6 \times 3) + (2 \times -1)$
$W = 18 - 2$
$W = 16 \text{ J}$.

(C) The work done $W$ in raising a box against gravity is given by the formula $W = mgh$,where $m$ is the mass of the box,$g$ is the acceleration due to gravity,and $h$ is the height by which it is raised.
Since $mg$ represents the weight of the box,the work done is equal to the product of the weight of the box and the height by which it is raised.
Therefore,the work done depends on the height $h$ to which the box is raised.

(D) The work done by a conservative force like gravity depends only on the initial and final vertical positions.
Let the lowest point be the reference level $(h=0)$.
At one extreme end,the height is $h$,so the potential energy is $U_i = mgh$.
At the other extreme end,the height is also $h$,so the potential energy is $U_f = mgh$.
The work done by the gravitational force is given by $W_g = -(U_f - U_i)$.
$W_g = -(mgh - mgh) = 0$.
Alternatively,the net vertical displacement of the bob from one extreme to the other is zero,and since gravity is a vertical force,the total work done is zero.

(C) The work done by the man is against gravity to lift his own weight and the load to a certain height $h$.
The formula for work done is $W = mgh$,where $m$ is the total mass $(50 \ kg + 10 \ kg = 60 \ kg)$,$g$ is the acceleration due to gravity,and $h$ is the height of the building.
Since the mass,gravity,and height remain constant regardless of the time taken to climb the building,the work done remains the same.
Therefore,the work done in $2 \ minutes$ is also $6 \times 10^4 \ J$.

(C) The formula for work done is given by $W = FS \cos \theta$.
Given: Force $F = 5 \ N$,Displacement $S = 10 \ m$,and Work $W = 25 \ J$.
Substituting these values into the formula:
$25 = 5 \times 10 \times \cos \theta$
$25 = 50 \times \cos \theta$
$\cos \theta = \frac{25}{50} = \frac{1}{2}$
Since $\cos 60^\circ = \frac{1}{2}$,the angle $\theta = 60^\circ$.

(B) In uniform circular motion,the centripetal force $F = \frac{mV^2}{r}$ is always directed towards the center of the circle.
At any given instant,the displacement $ds$ of the body is along the tangent to the circular path.
The angle between the centripetal force (directed towards the center) and the displacement (along the tangent) is always $90^\circ$.
The work done $W$ is given by the integral $W = \int F \cdot ds = \int F \cos(90^\circ) ds$.
Since $\cos(90^\circ) = 0$,the work done by the centripetal force is $0$ for any displacement,including half the circumference.

(C) The work done $W$ is defined as the dot product of force $\vec{F}$ and displacement $\vec{S}$:
$W = \vec{F} \cdot \vec{S} = F S \cos \theta$
where $\theta$ is the angle between the force and displacement vectors.
For the work done to be zero $(W = 0)$,we must have $\cos \theta = 0$.
This occurs when $\theta = 90^{\circ}$.
Therefore,the force $\vec{F}$ and displacement $\vec{S}$ must be mutually perpendicular to each other.

(A) The centripetal force $F$ acting on a particle moving in a circular path is always directed towards the center of the circle.
At any point on the circular path,the displacement vector $d\vec{s}$ is tangential to the path.
Since the radius vector is perpendicular to the tangent at any point,the centripetal force $F$ is always perpendicular to the instantaneous displacement $d\vec{s}$.
The work done $W$ is given by the integral $W = \int \vec{F} \cdot d\vec{s}$.
Since $\vec{F} \perp d\vec{s}$,the dot product $\vec{F} \cdot d\vec{s} = F \cdot ds \cdot \cos(90^{\circ}) = 0$.
Therefore,the total work done $W = \int 0 = 0$.

(A) The total work done by the man is the sum of work done in the horizontal and vertical directions.
Work done in the horizontal direction $(W_{Hz})$: Since the force (weight of the bucket) acts vertically downwards and the displacement is horizontal,the angle $\theta = 90^{\circ}$. Thus,$W_{Hz} = F \cdot d \cdot \cos(90^{\circ}) = 0 \ J$.
Work done in the vertical direction $(W_{vert})$: The man lifts the bucket against gravity,so the force applied is equal to the weight $(60 \ N)$ and the displacement is $5 \ m$ upwards. Thus,$W_{vert} = F \cdot d \cdot \cos(0^{\circ}) = 60 \ N \times 5 \ m = 300 \ J$.
Total work done $W_{net} = W_{Hz} + W_{vert} = 0 \ J + 300 \ J = 300 \ J$.

(D) Given: Mass $m = 20 \ kg$,Force $F = 5 \ N$,Initial velocity $u = 0$,Time $t = 1 \ s$.
First,calculate the acceleration $a$ using Newton's second law: $a = \frac{F}{m} = \frac{5}{20} = 0.25 \ m/s^2$.
Next,calculate the displacement $s$ in the first second using the equation of motion: $s = ut + \frac{1}{2}at^2 = 0(1) + \frac{1}{2}(0.25)(1)^2 = 0.125 \ m$.
Finally,calculate the work done $W$ using the formula $W = F \times s$: $W = 5 \times 0.125 = 0.625 \ J$.
Since $0.625 = \frac{5}{8}$,the work done is $\frac{5}{8} \ J$.

(C) When a ball is dropped from the top of a tower,the initial velocity $u = 0$.
The distance covered in the $n^{th}$ second is given by $h_n = u + \frac{g}{2}(2n - 1)$.
Since $u = 0$,$h_n \propto (2n - 1)$.
For $n = 1, 2, 3$,the distances are $h_1 : h_2 : h_3 = (2(1)-1) : (2(2)-1) : (2(3)-1) = 1 : 3 : 5$.
The work done by gravity is $W = mgh$.
Since $m$ and $g$ are constant,the ratio of work done is equal to the ratio of distances covered:
$W_1 : W_2 : W_3 = h_1 : h_2 : h_3 = 1 : 3 : 5$.

(B) The work done $W$ by a constant force is given by the dot product of the force vector $\vec{F}$ and the displacement vector $\vec{S}$.
Given: $\vec{F} = (-2\hat{i} + 15\hat{j} + 6\hat{k}) \, N$.
The displacement is along the $Y$-axis by $10 \, m$,so $\vec{S} = 10\hat{j} \, m$.
Work $W = \vec{F} \cdot \vec{S} = (-2\hat{i} + 15\hat{j} + 6\hat{k}) \cdot (10\hat{j})$.
Using the dot product property $\hat{i} \cdot \hat{j} = 0$,$\hat{j} \cdot \hat{j} = 1$,and $\hat{k} \cdot \hat{j} = 0$:
$W = (-2 \times 0) + (15 \times 10) + (6 \times 0) = 150 \, J$.

(B) Given: Force $F = 5 \ N$,mass $m = 15 \ kg$,initial velocity $u = 0$,time $t = 1 \ s$.
Using Newton's second law,acceleration $a = F/m = 5/15 = 1/3 \ m/s^2$.
The displacement $s$ in time $t$ is given by $s = ut + 1/2 at^2$.
Since $u = 0$,$s = 1/2 \times (1/3) \times (1)^2 = 1/6 \ m$.
The work done $W = F \times s = 5 \times (1/6) = 5/6 \ J$.

(B) The displacement vector $\vec{d}$ is given by the difference between the final position vector $\vec{r_2}$ and the initial position vector $\vec{r_1}$.
$\vec{d} = \vec{r_2} - \vec{r_1} = (8\hat i + 6\hat j - 3\hat k) - (4\hat i - 2\hat j + 5\hat k) = (4\hat i + 8\hat j - 8\hat k) \, m$.
Work done $W$ is the dot product of the force vector $\vec{F}$ and the displacement vector $\vec{d}$.
$W = \vec{F} \cdot \vec{d} = (5\hat i + 6\hat j - 7\hat k) \cdot (4\hat i + 8\hat j - 8\hat k)$.
$W = (5 \times 4) + (6 \times 8) + (-7 \times -8) = 20 + 48 + 56 = 124 \, J$.

(D) The work done $W$ by a force $\vec F$ for a displacement $\vec d$ is given by the dot product: $W = \vec F \cdot \vec d$.
Given that the work done is zero,we have $W = 0$.
Substituting the given vectors: $(6\hat i + 2\hat j - 3\hat k) \cdot (2\hat i - 3\hat j + c\hat k) = 0$.
Calculating the dot product: $(6 \times 2) + (2 \times -3) + (-3 \times c) = 0$.
$12 - 6 - 3c = 0$.
$6 - 3c = 0$.
$3c = 6$.
Therefore,$c = 2$.

(B) The maximum height reached by the particle is given by $h = \frac{u^2}{2g}$.
Substituting the values,$h = \frac{5^2}{2g} = \frac{25}{2g}$.
The work done by the force of gravity when the particle moves up is $W_{up} = -mgh = -m \cdot g \cdot \frac{25}{2g} = -1.25 \ J$.
When the particle returns to the starting point,the displacement is zero.
However,the question asks for the work done by gravity during the entire trip (up and down).
Work done by gravity during the upward journey is $W_{up} = -mgh = -1.25 \ J$.
Work done by gravity during the downward journey is $W_{down} = +mgh = +1.25 \ J$.
The total work done by gravity for the complete round trip is $W_{total} = W_{up} + W_{down} = -1.25 + 1.25 = 0 \ J$.
However,if the question implies the work done during the upward journey until it reaches the maximum height,the answer is $-1.25 \ J$.