Work, Energy and Power
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A body of man 100 kg travelled 10 m before coming to rest. If m = 0.4, work done against friction is (motion is happening on horizontal surface, take g=10 m/s2)
A body of man 100 kg travelled 10 m before coming to rest. If m = 0.4, work done against friction is (motion is happening on horizontal surface, take g=10 m/s2)
v2 = 2 × mgs
v2 = 2 × (0.4) × 10 × 10
v2 = 80
Wf = Dk
=
Wf = – 4000
Upvote 
A ball is released from a height of 1 m on a smooth hemispherical surface as shown. Find its velocity when it is at a height of 0.5 m. (Take g = 10 m/s2)
A ball is released from a height of 1 m on a smooth hemispherical surface as shown. Find its velocity when it is at a height of 0.5 m. (Take g = 10 m/s2)
By conservation of mechanical energy
mg (1) = mv2 + mg (0.5)
v2 = 10
v = m/s
Upvote Force on a particle moving in straight line is given by <!-- [if gte mso 9]><xml>
Force on a particle moving in straight line is given by <!-- [if gte mso 9]><xml>
= 18t4 – 18t2
->P (t = 2) =18 [16 − 4] = 216 W
Upvote If the coefficient of restitution is 0.5, what is the percentage loss of energy on each rebounding of a ball dropped from a height?
If the coefficient of restitution is 0.5, what is the percentage loss of energy on each rebounding of a ball dropped from a height?
h2 = e2h1
h2 = h1/4
Energy loss =mgh1 - mgh2 /mgh1*100 = 75%
Upvote The kinetic energy of a body decreases by 36%. The decrease in linear momentum is
The kinetic energy of a body decreases by 36%. The decrease in linear momentum is
A particle of mass m is dropped from rest from a height h0 on a horizontal floor. The coefficient of restitution between the floor and particle is e. Maximum height attained by the particle after second collision is
A particle of mass m is dropped from rest from a height h0 on a horizontal floor. The coefficient of restitution between the floor and particle is e. Maximum height attained by the particle after second collision is
Height attained after first collision (h1) = e2h0
h2 = e2h1 = e4h0
Upvote The potential energy of a particle varies with x according to the relation U(x) = x2 – 4x. The point x = 2 is a point of
The potential energy of a particle varies with x according to the relation U(x) = x2 – 4x. The point x = 2 is a point of
A small block of mass 10 g is projected along the smooth path from rest as shown in the given figure below. Find its speed at the point B
<!--[endif]--><!-- [if gte mso 9]><xml>
A small block of mass 10 g is projected along the smooth path from rest as shown in the given figure below. Find its speed at the point B
<!--[endif]--><!-- [if gte mso 9]><xml>
m
h=
By work energy theorem
gh
Upvote