Category: Work Energy & Power

Q: A block of mass 2 kg is free to move along the x-axis, it is at rest and from t=0 onwards it is subjected to a time-dependent force F(t) in the x-direction. The force F(t) varies with t as shown in the figure. The kinetic energy of the block after 4.5 s is

(a)4.50 J

(b)7.50 J

(c)5.06 J

(d)14.06 J

Ans: (c)

Sol: Change in Momentum p = Area of F-t graph

$\large p = \frac{1}{2}(4\times 3) – \frac{1}{2}(1.5 \times 2)$

Kinetic Energy $\large K = \frac{p^2}{2 m}$

$\large K = \frac{(\frac{1}{2}(4\times 3) – \frac{1}{2}(1.5 \times 2))^2}{2 \times 2}$

= 5.06 J

Q: A bob of mass M is suspended by a massless string of length L the horizontal velocity v at position A is just sufficient to make it reach the point B. The angle θ at which the speed of the bob is half of that at A, satisfies

(a) θ = π/4

(b)π/4 < θ<π/2

(c)π/2 < θ = 3π/4

(d)3π/4 < θ  < π

Ans: (d)

Sol: As velocity at A is

$\large v = \sqrt{5 g L}$ …(i)

Applying Conservation of Energy between point A & other point (when particle makes an angle θ);

$\large \frac{1}{2}m v^2 = \frac{1}{2}m (v/2)^2 + m g h$

$\large v^2 = (v/2)^2 + 2 g h$

$\large v^2 = \frac{v^2}{4} + 2 g L(1-cos\theta)$

$\large \frac{3 v^2}{4} = 2 g L(1-cos\theta)$

$\large \frac{3 \times 5 g L}{4} = 2 g L(1-cos\theta)$

$\large \frac{15}{8} = 1-cos\theta $

$\large -\frac{7}{8} = cos\theta $

$\large \theta = cos^{-1}(-\frac{7}{8})$

θ = 151°

Q: If W1,W2 and W3 represent the work done In moving a particle from A to B along three different paths 1,2 and 3 respectively(as shown) in the gravitational field of a point mass m. Find the correct relation between W1,W2 and W3

(a) W1 > W2 > W3

(b) W1 = W2 = W3

(c) W1< W2 <W3

(d) W2>W1>W3

Ans: (b)

Sol: Gravitational field is a conservative force field . In a conservative force force field work done is independent of path .

Hence , W1 = W2 = W3

Q: When a rubber band is stretched by a distance x, it exerts a restoring force of magnitude F= ax+bx², where a and b are constants. The work done in stretching the unstretched rubber-band by L is

(a)$\large (a L^2 + b L^3)$

(b)$\large \frac{1}{2}(a L^2 + b L^3)$

(c)$\large (\frac{a L^2}{2} + \frac{b L^3}{3} )$

(d)$\large \frac{1}{2}(\frac{a L^2}{2} + \frac{b L^3}{3} )$

Ans: (d)

Sol: Work done by stretching the rubber band by L is

$\large W = \int_{0}^{L} (ax + bx^2 ) dx$

$\large W = [\frac{ax^2}{2}]_{0}^{L} + [\frac{bx^3}{3}]_{0}^{L}$

$\large W = [\frac{aL^2}{2}] + [\frac{bL^3}{3}]$

Q: The work done on a particle of mass m by a force, $\large K[\frac{x}{(x^2 + y^2)^{3/2}}\hat{i} + \frac{y}{(x^2 + y^2)^{3/2}}\hat{j}]$ (K being a constant appropriate dimensions), when the particle is taken from the point (a, 0) to the point (0, a) along a circular path of radius a about the origin in the x-y plane is

(a)(2 Kπ)/a

(b)( Kπ)/a

(c)( Kπ)/(2 a)

(d)0

Ans: (d)

Sol: Let P(x,y) be the point on circular path

Position vecrot $ \vec{r}=\vec{OP} = x \hat{i} + y \hat{j}$

$\vec{F} = \frac{K}{(x^2 + y^2)^{3/2}}(x \hat{i} + y \hat{j}) = \frac{K}{r^3}\vec{r}$

Since F is along radial direction , therefore work done is zero.