A block of mass m is sliding down an inclined plane with constant spee

1.	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» 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 ⇒ W_friction + W_gravitation = ∆KE ⇒ W_friction= - W_gravitation or W_friction = - mgh = So, energy dissipated due to friction = mgh.

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»

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

⇒ W_friction + W_gravitation = ∆KE

⇒ W_friction= - W_gravitation

or W_friction = - mgh

= So, energy dissipated due to friction = mgh.

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