# 23 - 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

**A particle is moving in a circular path of radius r under the action of a force F. If at an instant velocity of particle is v, and speed of particle is increasing, then **

**(i) F.v=0 (ii) F.v > 0 (iii) F.v < 0**

**Asked By: SWATI KAPOOR**

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**Joshi sir comment**

centripetal acc. is the need for circular motion but speed is increasing so tangential acc will also be present so angle between velocity and acc will be acute so **F.v > 0**

**A ball of mass m moving with speed v collides with a smooth horizontal surface at angle θ with it as shown in figure. The magnitude of impulse imparted to surface by ball is **

**Asked By: SWATI KAPOOR**

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**Joshi sir comment**

From where θ is measured, horizontal or vertical ?

A particle of mass m moving from origin along x-axis and its velocity varies with position x as v= k√x. The work done by force acting on it during first 't' seconds is

**Asked By: SWATI KAPOOR**

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**Joshi sir comment**

v = k√x

so a = dv/dt = [k/2√x]*dx/dt = [k/2√x]*v = [k/2√x]*[k√x] = k^{2}/2 = constant

and dx/dt = k√x

so dx/√x = kdt

integrate within o to t

finally work = Fx because F is constant

**A particle of mass m is projected with speed u at angle θ with horizontal from ground. The work done by gravity on it during its upward motion is**

**Asked By: SWATI KAPOOR**

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**Joshi sir comment**

H = u^{2}sin^{2}θ/2g

so work done by gravity = mgH

**A body of mass m is projected from ground with speed u at an angle θ with the horizontal. The power delivered by gravity to it at half of maximum height from the ground is **

**Asked By: SWATI KAPOOR**

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**Joshi sir comment**

at half height, vertical velocity v^{2} = u^{2}sin^{2}θ - 2gH/2, from this equation calculate v

then P = F.v = -mg* vertical component of velocity

this will be instataneous power