# 28 - Work, Energy and Power Questions Answers

a block of mass m is pushed up a movable incline of mass nm and height h.All surfaces are smooth without friction.what must be the minimum value of u so that the block just reaches the top of the movable incline.

**Asked By: AD**

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**Joshi sir comment**

let velocity of m at start is u along horizontal

at the top horizontal velocity of system = u/(n+1) and vertical velocity = 0

by energy conservation 1/2 mu^{2} = 1/2 (nm+m) {u/(n+1)}^{2} + mgh

solve

A block of mass 100 g is moved with a speed of 5 m/s at the highest point in a closed circular tube of radius 10 cm kept in vrtical plane.The cross section of the tube is such that the block just fits in it.The block makes several oscillations inside the tube and finally stops at the lowest point. Find the work done by the tube on the block.

**Asked By: SUBHASH**

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**Joshi sir comment**

Total energy is lost so K E will be the work done

**The rate of doing work by force acting on a particle moving along x-axis depends on position x of particle and is equal to 2x. The velocity of particle is given by **

**Asked By: SWATI KAPOOR**

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**Joshi sir comment**

Rate of doing work is power so according to the given condition

P = 2x

so Fv = 2x

or m(dv/dt)v = 2x

or m(vdv/dx)v = 2x

or mv^{2}dv = 2xdx

now integrate.

**A shell at rest on a smooth horizontal surface explodes into two fragments of masses m1 and m2. If just after explosion m1 moves with speed u, then work done by internal forces during explosion is**

**Asked By: SWATI KAPOOR**

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**Joshi sir comment**

By momentum conservation velocity of m_{2} = m_{1}u/m_{2}

now sum of kinetic energies = Net work done by internal forces

**A body of mass m falls from height h on ground. If 'e' be the coefficient of restitution of collision between the body and ground, then the distance travelled by body before it comes to rest is**

**Asked By: SWATI KAPOOR**

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**Joshi sir comment**

if a body falls from height h then height attained after rebound will be e^{2}h

by using this fact

total distance covered before stop = h+2e^{2}h+2e^{2}e^{2}h+2e^{2}e^{2}e^{2}h+..................................................

so d = h+2e^{2}h[1+e^{2}+e^{4}+e^{6}+.........................]

now solve