Work Done by Constant and Variable Forces

Practice Questions:

Question 31: easy

A body moves a distance of $10\text{ m} $ along a straight line under the action of a $5\text{ N}$ force. If the work done is $25\text{ J}$, then the angle between the force and direction of motion of the body is

(1997)

1. $60^{\circ}$

2. $75^{\circ}$

3. $30^{\circ}$

4. $45^{\circ}$

View Answer

Work $W = Fd cos\theta$. Given $W = 25\text{ J}$, $F = 5\text{ N}$, $d = 10\text{ m})$. $25 = (5)(10) cos\theta$. $cos\theta = \frac{1}{2}$. Therefore, $\theta = 60^{\circ}$.

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Question 32: moderate

A body, constrained to move in $y$-direction, is subjected to a force given by $F = (-2\hat{i} + 15\hat{j} + 6\hat{k})\text{ N})$. The work done by this force in moving the body through a distance of $10\hat{j}\text{ m})$ along $y$-axis, is:

(1994)

1. 150 J

2. 20 J

3. 190 J

4. 160 J

View Answer

Force $F = (-2hat{i} + 15hat{j} + 6hat{k})$. Displacement $dr = 10hat{j}\text{ m})$. Work $W = F cdot dr$. $W = (15)(10) = 150\text{ J}$.

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Question 33: moderate

A block of mass $10\text{ kg}$ moving in $x$ direction with a constant speed of $10\text{ m s}^{-1}$, is subjected to a retarding force $F = -0.1x\text{ J/m}$ during its travel from $x = 20\text{ m}$ to $30\text{ m}$. Its final K.E. will be: (2015)

1. $450\text{ J}$

2. $275\text{ J}$

3. $250\text{ J}$

4. $475\text{ J}$

View Answer

Initial kinetic energy is $K_i = \frac{1}{2}mv^2 = 500\text{ J}$. Work done by the retarding force is $W = \int_{20}^{30} (-0.1x)dx = -25\text{ J}$. Using work-energy theorem, final K.E. $K_f = K_i + W = 500 - 25 = 475\text{ J}$.

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