Work power energy is the most important chapter when it comes to mechanics, for JEE (Advanced). The chapter is quite tricky and takes a lot of time and devotion on your part to understand and master. But this chapter is a complete gold medal for almost all the questions of the mechanics can be solved by the work power energy approach if you master this topic..

**Q1.**

Power supplied to a particle of mass 2 kg varies with time as P=3t^2/2 W. Here t is in second. If the velocity of particle at t=0 is v=0, the velocity of particle at time t=2 s will be

**Q2.**

The speed v reached by a car of mass m in travelling a distance x, driven with constant power P, is given by

Solution

From work-energy theorem,

W=∆KE=K_f-K_i=1/2 m(v_f^2-v_i^2 )

=1/2×2[(-20)^2-(10)^2 ]=300 J

W=∆KE=K_f-K_i=1/2 m(v_f^2-v_i^2 )

=1/2×2[(-20)^2-(10)^2 ]=300 J

Solution

Let x be the extension in the string when 2 kg block leaves the contact with ground. Then tension in the spring should be equal to weight of 2 kg block:

Kx=2g or x=2g/K=(2×10)/40=1/2m

Now from conservation of mechanical energy,

mgx=1/2 Kx^2+1/2 mv^2

or v=√(2gx-(Kx^2)/m)=√(2×10×1/2-40/(4×5))=2√2 ms^(-1)

Kx=2g or x=2g/K=(2×10)/40=1/2m

Now from conservation of mechanical energy,

mgx=1/2 Kx^2+1/2 mv^2

or v=√(2gx-(Kx^2)/m)=√(2×10×1/2-40/(4×5))=2√2 ms^(-1)

Solution

Work done by man, mgL=mg(L-s)+mgs where mg(L-s) is the increase in potential energy of the man and mgs is the increase in potential energy of the balloon because the4 balloon would have been lifted up but for the climbing of the man. So
(Increase in PE og man)/(Increase in PE of balloon)=(mg(L-s))/mgs=(L-s)/s

**Q7.**

An object of mass m slides down a hill of orbitrary shape and after travelling a certain horizontal path stops because of friction. The total vertical height descended ish. The friction coefficient is different for different segments for the entire path but is independent of the velocity and direction of motion. The work that a tangential force must perform to return the object to its initial position along the same path is

Solution

The initial potential energy of the object is mgh, which has been used up for work against the force of friction. In returning the body to its initial position, the force performs the same work against friction. In addition, it imparts top the object the initial potential energy. As a result, the total work will be 2mgh

**Q8.**

**Statement 1**: If two protons are brought near one another, the potential energy of the system will increase

**Statement 2:**The change on the proton is +1.6×10^(-19) C

Solution

If two protons are brought near one another, work has to be done against electrostatic force because same charge repel each other. The work done is stored as potential energy in the system

**Q9.**

**Statement 1**: A spring has potential energy, both when it is compressed or stretched

**Statement 2:**In compressing or stretching, work is done on the spring against the restoring force

Solution

The work done on the spring against the restoring force is stored as potential energy in both conditions when it is compressed or stretched

**Q10. Statement 1:**If a particle of mass m is connected to a light rod and whirled is a vertical circle of radius R, then to complete the circle, the minimum velocity of the particle at the bottom point is √5gR

**Statement 2:**Mechanical energy is conserved and in case of the minimum velocity at the bottom point, the velocity at the top point will be zero

Solution

Using conservation of mechanical energy and taking zero velocity at the bottom point is 2√gR