## A block of weight W is dragged across the horizontal floor from A to B by the constant vertical force ….

Q: A block of weight W is dragged across the horizontal floor from A to B by the constant vertical force P acting at the end of the rope. Calculate the work done on the block by the force P = ( √3 + 1)N.Assume that block does not lift off the floor. (g = 10 m/s2 )

Click to See Solution :
Ans: 4 J
Sol: W = P x

$W = P[\frac{h}{sin30^o} – \frac{h}{sin60^o} ]$

$W = P[\frac{h}{1/2} – \frac{h}{\sqrt{3}/2} ]$

$W = 2Ph[1 – \frac{1}{\sqrt{3}} ]$

$W = 2Ph[\frac{\sqrt{3}-1}{\sqrt{3}}]$

$W = 2(\sqrt{3}+1)\sqrt{3}[\frac{\sqrt{3}-1}{\sqrt{3}}]$

W = 4 J

## Power delivered to a body varies as P = 3 t^2 . Find out the change in kinetic energy of the body …

Q: Power delivered to a body varies as P = 3 t2 . Find out the change in kinetic energy of the body from t = 2 to t = 4 sec.

Click to See Solution :
Ans: 56 J
Sol: Applying work energy theorem to body ΔKE = work done by forces delivering power

$= \int_{t=2}^{t=4} P dt$

$= \int_{t=2}^{t=4} 3 t^2 dt$

= 56 J

## The potential energy of a particle varies with x according to the relation U(x) = x^2 – 4 x. The point x = 2 is a point of ….

Q: The potential energy of a particle varies with x according to the relation U(x) = x2 – 4 x. The point x = 2 is a point of :

(A) stable equilibrium

(B) unstable equilibrium

(C) neutral equilibrium

(D) none

Click to See Solution :
Ans: (A)
Sol: U(x) = x2 – 4x

F = 0

$\frac{dU(x)}{dx} = 0$

2 x – 4 = 0

x = 2

$\frac{d^2U(x)}{dx^2} = 2 > 0$

i.e. U is minimum hence x = 2 is a point of stable
equilibrium

## A block is attached with a spring and is moving towards a fixed wall with speed v as shown in figure….

Q: A block is attached with a spring and is moving towards a fixed wall with speed v as shown in figure. As the spring reaches the wall, it starts compressing. The work done by the spring on the wall during the process of compression is :

(A) $\frac{1}{2} m v^2$

(B) $m v^2$

(C) Kmv

(D) zero

Click to See Solution :
Ans: (D)
Sol: As point of application of force is not moving,
therefore work done by the force is zero