28 - Work, Energy and Power Questions Answers

Dear Sir request if you can help provide solution 
Asked By: RK GUPTA
is this question helpfull: 0 0 submit your answer
Dear Sir request if you can help provide solution 
Asked By: RK GUPTA
is this question helpfull: 0 0 submit your answer
Dear Sir request if you can help provide solution 
Asked By: RK GUPTA
is this question helpfull: 0 0 submit your answer
Asked By: LALIT JOSHI
is this question helpfull: 0 0 submit your answer
Joshi sir comment
Asked By: MITUL
is this question helpfull: 0 0 submit your answer
Joshi sir comment

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
is this question helpfull: 2 2 submit your answer
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 mu2 = 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
is this question helpfull: 4 5 submit your answer
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
is this question helpfull: 5 1 read solutions ( 1 ) | submit your answer
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 mv2dv = 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
is this question helpfull: 5 6 submit your answer
Joshi sir comment

By momentum conservation velocity of m2 = m1u/m2

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
is this question helpfull: 4 7 submit your answer
Joshi sir comment

if a body falls from height h then height attained after rebound will be e2h

by using this fact

total distance covered before stop = h+2e2h+2e2e2h+2e2e2e2h+..................................................

so d = h+2e2h[1+e2+e4+e6+.........................]

now solve 

 

Login Here

Register