Question 31:
easyA 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)
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, \(\theta = 60^{\circ}\).
Question 32:
moderateA 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)
Force \(F = (-2hat{i} + 15hat{j} + 6hat{k})\). Displacement \(dr = 10hat{j}\text{ m})\). Work \(W = F cdot dr\). \(W = (15)(10) = 150\text{ J}\).
Question 33:
moderateA 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)
Use work-energy theorem: $K_f - K_i = W$. Initial kinetic energy is $K_i = \frac{1}{2}mv^2 = 500\text{ J}$. Work done by the force is $W = \int_{20}^{30} (-0.1x) dx = -25\text{ J}$. Thus, final kinetic energy is $K_f = 500 - 25 = 475\text{ J}$.
Question 34:
easy$300\text{ J}$ of work is done in sliding a $2\text{ kg}$ block up an inclined plane of height $10\text{ m}$. Taking $g = 10\text{ m/s}^2$, work done against friction is:
(2006)
Total work done on the block goes into increasing potential energy and overcoming friction. Potential energy gain is $U = mgh = 2 \times 10 \times 10 = 200\text{ J}$. Work done against friction = Total work - $U = 300 - 200 = 100\text{ J}$.