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A block of mass m is sliding down an inclined plane with constant speed. At a certain instant t0, its height above the ground is h. The coefficient of kinetic friction between the block and the plane is µ. If the block reaches the ground at a later instant tg, then the energy dissipated by friction in the time interval (tg – t0) is
(a) µmgh
(b) µmgh/sinθ
(c) mgh
(d) µmgh/cosθ
A block of mass m is sliding down an inclined plane with constant speed. At a certain instant t0, its height above the ground is h. The coefficient of kinetic friction between the block and the plane is µ. If the block reaches the ground at a later instant tg, then the energy dissipated by friction in the time interval (tg – t0) is
(a) µmgh
(b) µmgh/sinθ
(c) mgh
(d) µmgh/cosθ
Answer is : (c) mgh
As block is sliding with a constant speed, so change in kinetic energy of block when it reaches bottom is zero.
Now, by work-energy theorem,
Total work done = Change in kinetic energy
⇒ Wfriction + Wgravitation = ∆KE
⇒ Wfriction = – Wgravitation
or Wfriction = – mgh
= So, energy dissipated due to friction = mgh.