## A spring placed horizontally on a rough horizontal surface is compressed against a block of mass m placed on the surface so as to store maximum energy in the spring. If the coefficient of friction between the block and the surface is μ , the potential energy stored in the spring is

Q: A spring placed horizontally on a rough horizontal surface is compressed against a block of mass m placed on the surface so as to store maximum energy in the spring. If the coefficient of friction between the block and the surface is μ , the potential energy stored in the spring is
(A) μ2m2g2/2k
(B) 2μm2g2/k
(C) μ2m2g/2k
(D) 3μ2mg2/k

Sol: For equilibrium of the block
Fmax – μmg = 0
 F = μmg
 U = F2/2k
U = μ2m2g2/2k

## A particle moves with a velocity 5i-3j+6k m/s under the influence of a constant force F= 10i+10j+20k . The instantaneous power applied to the particle is

Q:A particle moves with a velocity m/s under the influence of a constant force . The instantaneous power applied to the particle is:
(A) 200 J/s
(B) 40 J/s
(C) 140 J/s
(D) 170 J/s

Solution : P = F.v
= (10i + 10j + 20k ).(5i – 3j + 6k )
= 50 – 30 + 120
= 140 J/s

## A block of mass m is moving with a constant acceleration a on a rough horizontal plane. If the coefficient of friction between the block and ground is μ, the power delivered by the external agent after a time t from the beginning is equal to

Q:A block of mass m is moving with a constant acceleration a on a rough horizontal plane. If the coefficient of friction between the block and ground is μ, the power delivered by the external agent after a time t from the beginning is equal to
(A) ma2t
(B) μmgat
(C) μm(a+μg)gt
(D ) m(a+μg)at
Sol: Instantaneous power delivered = P = F.v = F v
where, F – f = ma
 F = f + ma
 P = (f + ma) v
Put f = μmg
 P = ( μmg + ma)v = m(a + μg).at

## A pumping machine pumps a liquid at a rate of 60 cc per minute at a pressure of 1.5 atmosphere

A pumping machine pumps a liquid at a rate of 60 cc per minute at a pressure of 1.5 atmosphere. The power of the machine is
(A) 9 watts
(B) 6 watts
(C ) 9 kW
(D) None of these

Solution : Power = F.v , where F = force imparted by the machine , v = velocity of the liquid
P = p.A.v, Where p = pressure & A = effective area
= p dV/dt
= (1.5×105 N/m2)(60 ×10-6 N/m2)
( 1 atm ≈ 105 N/m2 )
= 9 watts

## When a man walks on a horizontal surface with constant velocity, work done by

When a man walks on a horizontal surface with constant velocity, work done by

(A) Friction is zero

(B) Contact force is zero

(C) Gravity is zero

(D) Man is zero

Solution :  Since mg & N are perpendicular to velocity and , work done by these forces is zero.   Since no relative sliding occurs during walking, static friction comes into play.  Hence the point of application of static frictional force does not move.  Therefore no work is done by frictional force.

The man has to lose his body’s (internal) energy E , hence performs  work because   W = ΔE