Work Done by Constant Force Questions in English

(C) When a man pushes a wall,he does not produce any displacement in the wall.
Since the displacement $s = 0$,the work done $W$ is calculated as:
$W = F \cdot s \cdot \cos(\theta)$
$W = F \cdot 0 \cdot \cos(\theta) = 0$
Therefore,the man does no work at all.

(D) The given force is $F = K \left[ \frac{x}{(x^2+y^2)^{3/2}} \hat{i} + \frac{y}{(x^2+y^2)^{3/2}} \hat{j} \right]$.
In polar coordinates,$x = r \cos \theta$ and $y = r \sin \theta$,where $r = \sqrt{x^2+y^2}$.
Substituting these into the force expression:
$F = K \left[ \frac{r \cos \theta}{r^3} \hat{i} + \frac{r \sin \theta}{r^3} \hat{j} \right] = \frac{K}{r^2} (\cos \theta \hat{i} + \sin \theta \hat{j})$.
This force is a central force directed radially outward,i.e.,$F = \frac{K}{r^2} \hat{r}$.
The work done by a central force is given by $W = \int F \cdot dr = \int \frac{K}{r^2} dr$.
Since the particle moves along a circular path of radius $a$,the distance $r$ from the origin is constant $(r = a)$.
For a circular path,the displacement vector $d\vec{l}$ is always perpendicular to the radial force vector $\vec{F}$.
Therefore,the dot product $\vec{F} \cdot d\vec{l} = 0$ at every point on the path.
Thus,the total work done $W = \int \vec{F} \cdot d\vec{l} = 0$.

(C) Given: Force $F = 10 \ N$,mass $m = 1.5 \ kg$,coefficient of kinetic friction $\mu_k = 0.2$,$g = 10 \ m/s^2$,initial velocity $u = 0$,time $t = 6 \ s$.
First,calculate the kinetic friction force: $f_k = \mu_k \cdot m \cdot g = 0.2 \times 1.5 \times 10 = 3 \ N$.
The net force acting on the block is $F_{net} = F - f_k = 10 - 3 = 7 \ N$.
The acceleration of the block is $a = F_{net} / m = 7 / 1.5 = 14 / 3 \ m/s^2$.
The displacement $s$ in $t = 6 \ s$ is given by $s = ut + (1/2)at^2 = 0 + (1/2) \times (14/3) \times (6)^2 = (1/2) \times (14/3) \times 36 = 7 \times 12 = 84 \ m$.
The work done by the applied force is $W = F \times s = 10 \times 84 = 840 \ J$.

(C) The displacement of the object is along the $X$-axis by $3 \text{ m}$, so the displacement vector is $\vec{d} = 3 \hat{i} \text{ m}$.
Given the force vector is $\vec{F} = (2 \hat{i} + 4 \hat{j}) \text{ N}$.
The work done $W$ by a constant force is given by the dot product of the force vector and the displacement vector:
$W = \vec{F} \cdot \vec{d}$
$W = (2 \hat{i} + 4 \hat{j}) \cdot (3 \hat{i})$
$W = (2 \times 3)(\hat{i} \cdot \hat{i}) + (4 \times 0)(\hat{j} \cdot \hat{i})$
Since $\hat{i} \cdot \hat{i} = 1$ and $\hat{j} \cdot \hat{i} = 0$, we get:
$W = 6 \times 1 + 0 = 6 \text{ J}$.

(D) Mass of the body, $m = 10 \,kg$. Force, $F = 4 \,N$. Acceleration produced in the body due to the applied force is $a = \frac{F}{m} = \frac{4}{10} = 0.4 \,m/s^2$.
For the interval $0 \leq t \leq 1 \,s$, the distance traveled $s_1$ is $s_1 = u t + \frac{1}{2} a t^2 = 0 + \frac{1}{2} \times 0.4 \times (1)^2 = 0.2 \,m$.
Work done $W_1 = F \times s_1 = 4 \times 0.2 = 0.8 \,J$.
For the interval $1 \,s \leq t \leq 2 \,s$, the initial velocity $v_i$ at $t = 1 \,s$ is $v_i = u + a t = 0 + 0.4 \times 1 = 0.4 \,m/s$.
The distance traveled $s_2$ during this interval $(t = 1 \,s)$ is $s_2 = v_i t + \frac{1}{2} a t^2 = 0.4 \times 1 + \frac{1}{2} \times 0.4 \times (1)^2 = 0.4 + 0.2 = 0.6 \,m$.
Work done $W_2 = F \times s_2 = 4 \times 0.6 = 2.4 \,J$.
The ratio $\frac{W_2}{W_1} = \frac{2.4}{0.8} = 3$.

(A) Given,mass of particle,$m = 30 \,g = 3 \times 10^{-2} \,kg$.
The position of the particle is $x = \alpha t^2$.
Velocity $v = \frac{dx}{dt} = 2\alpha t$.
Acceleration $a = \frac{dv}{dt} = 2\alpha$.
Force $F = ma = (3 \times 10^{-2} \,kg) \times (2 \times 1 \,m/s^2) = 6 \times 10^{-2} \,N$.
Work done $W = \int F dx$. Since $x = \alpha t^2$,$dx = 2\alpha t dt$.
At $t = 0, x = 0$. At $t = 4, x = 1 \times (4)^2 = 16 \,m$.
$W = \int_{0}^{16} (6 \times 10^{-2}) dx = 6 \times 10^{-2} \times [x]_{0}^{16} = 6 \times 10^{-2} \times 16 = 0.96 \,J$.

(A) Given:
$F_{1} = (\hat{i} + 2\hat{j} + 3\hat{k}) \text{ N}$
$F_{2} = (4\hat{i} - 5\hat{j} - 2\hat{k}) \text{ N}$
$r_{1} = (20\hat{i} + 15\hat{j}) \text{ cm} = (0.2\hat{i} + 0.15\hat{j}) \text{ m}$
$r_{2} = (7\hat{k}) \text{ cm} = (0.07\hat{k}) \text{ m}$
Total force $F = F_{1} + F_{2} = (1+4)\hat{i} + (2-5)\hat{j} + (3-2)\hat{k} = (5\hat{i} - 3\hat{j} + \hat{k}) \text{ N}$.
Displacement $s = r_{2} - r_{1} = (0\hat{i} + 0\hat{j} + 0.07\hat{k}) - (0.2\hat{i} + 0.15\hat{j} + 0\hat{k}) = (-0.2\hat{i} - 0.15\hat{j} + 0.07\hat{k}) \text{ m}$.
Work done $W = F \cdot s = (5\hat{i} - 3\hat{j} + \hat{k}) \cdot (-0.2\hat{i} - 0.15\hat{j} + 0.07\hat{k})$.
$W = (5 \times -0.2) + (-3 \times -0.15) + (1 \times 0.07)$.
$W = -1.0 + 0.45 + 0.07 = -0.48 \text{ J}$.

(C) The initial position vector of the particle is $\vec{r}_1 = \hat{i} + 3 \hat{k}$.
The final position vector of the particle is $\vec{r}_2 = -3 \hat{i} + 4 \hat{j} + 5 \hat{k}$.
The force acting on the particle is $\vec{F} = \hat{i} + 5 \hat{k}$.
The displacement vector $\vec{s}$ is given by $\vec{r}_2 - \vec{r}_1$:
$\vec{s} = (-3 \hat{i} + 4 \hat{j} + 5 \hat{k}) - (\hat{i} + 3 \hat{k})$
$\vec{s} = -4 \hat{i} + 4 \hat{j} + 2 \hat{k}$.
Work done $W$ is the dot product of force and displacement:
$W = \vec{F} \cdot \vec{s} = (\hat{i} + 5 \hat{k}) \cdot (-4 \hat{i} + 4 \hat{j} + 2 \hat{k})$.
Calculating the dot product:
$W = (1)(-4) + (0)(4) + (5)(2)$
$W = -4 + 0 + 10 = 6 \ J$.

(D) Given: Mass $m = 6 \,kg$, Displacement $s = \frac{t^2}{4} \,m$.
Velocity $v = \frac{ds}{dt} = \frac{d}{dt}(\frac{t^2}{4}) = \frac{2t}{4} = \frac{t}{2} \,m/s$.
Acceleration $a = \frac{dv}{dt} = \frac{d}{dt}(\frac{t}{2}) = \frac{1}{2} \,m/s^2$.
Force $F = m \times a = 6 \times \frac{1}{2} = 3 \,N$.
At $t = 2 \,s$, displacement $s = \frac{(2)^2}{4} = \frac{4}{4} = 1 \,m$.
Work done $W = F \times s = 3 \,N \times 1 \,m = 3 \,J$.

(B) To pump the water out of the vessel, we need to lift the center of mass of the water to the top level of the vessel.
Let the side length of the cube be $L = 1 \,m$.
The volume of the water is $V = L^3 = 1^3 = 1 \,m^3$.
The density of water is $\rho = 1000 \,kg/m^3$.
The mass of the water is $m = \rho V = 1000 \times 1 = 1000 \,kg$.
The center of mass of the water in a full cubical vessel is at a height of $h_{cm} = L/2 = 0.5 \,m$ from the bottom.
To pump the water out, the center of mass must be raised to the top of the vessel, which is at a height $h = L/2 = 0.5 \,m$ from the initial center of mass position.
The work done is $W = mgh = 1000 \times 10 \times 0.5 = 5000 \,J$.

(D) The mass per unit length of the rope is $\lambda = \frac{M}{L} = \frac{0.4 \,kg}{4 \,m} = 0.1 \,kg/m$.
The length of the hanging part is $l = 0.6 \,m$.
The mass of the hanging part is $m = \lambda \times l = 0.1 \,kg/m \times 0.6 \,m = 0.06 \,kg$.
The center of mass of the hanging part is at a distance $h = \frac{l}{2} = \frac{0.6 \,m}{2} = 0.3 \,m$ below the table edge.
The work done to pull the rope onto the table is equal to the change in potential energy of the hanging part,which is $W = mgh$.
Substituting the values: $W = 0.06 \,kg \times 10 \,m/s^2 \times 0.3 \,m = 0.18 \,J$.

(C) Given: $m = 2 \ kg$,$x(t) = t^2 + t + 1 \ m$.
The velocity is $v(t) = \frac{dx}{dt} = 2t + 1 \ m/s$.
The acceleration is $a(t) = \frac{dv}{dt} = 2 \ m/s^2$.
Since the acceleration is constant,the force acting on the body is $F = m \times a = 2 \ kg \times 2 \ m/s^2 = 4 \ N$.
The displacement $s$ during the interval $t = 2 \ s$ to $t = 3 \ s$ is $s = x(3) - x(2)$.
$x(3) = (3)^2 + 3 + 1 = 9 + 3 + 1 = 13 \ m$.
$x(2) = (2)^2 + 2 + 1 = 4 + 2 + 1 = 7 \ m$.
$s = 13 - 7 = 6 \ m$.
The work done is $W = F \times s = 4 \ N \times 6 \ m = 24 \ J$.

(C) The resultant force $\vec{F} = \vec{F}_1 + \vec{F}_2 = (2+3)\hat{i} + (3-1)\hat{j} + (4-2)\hat{k} = (5\hat{i} + 2\hat{j} + 2\hat{k})\text{ N}$.
The displacement vector $\vec{d}$ is along the direction $(3\hat{i} - 4\hat{j})$.
The unit vector in this direction is $\hat{u} = \frac{3\hat{i} - 4\hat{j}}{\sqrt{3^2 + (-4)^2}} = \frac{3\hat{i} - 4\hat{j}}{5}$.
Thus,the displacement vector is $\vec{d} = 25 \hat{u} = 25 \times \frac{3\hat{i} - 4\hat{j}}{5} = 5(3\hat{i} - 4\hat{j}) = (15\hat{i} - 20\hat{j})\text{ m}$.
The work done $W$ is given by the dot product of force and displacement: $W = \vec{F} \cdot \vec{d} = (5\hat{i} + 2\hat{j} + 2\hat{k}) \cdot (15\hat{i} - 20\hat{j} + 0\hat{k})$.
$W = (5 \times 15) + (2 \times -20) + (2 \times 0) = 75 - 40 = 35\text{ J}$.