Question

What If? The block of mass m 5200 g described in Problem 43 (Fig. P8.43) is released from rest at point A, and the surface of the bowl is rough. The block’s speed at point B is 1.50 m/s. (a) What is its kinetic energy at point B? (b) How much mechanical energy is transformed into internal energy as the block moves from point A to point B? (c) Is it possible to determine the coefficient of friction from these results in any simple manner? (d) Explain your answer to part (c).

Short answer

(a) Its kinetic energy at point B is.

(b) The mechanical energy is transformed into internal energy as the block moves from point A to point B is.

(c) No, we cannot determine the coefficient.

(d) As normal reaction on the box and frictional force both vary with position, it is not possible to find the value of coefficient of friction.

Step-by-step solution

Step-by-Step SolutionStep 1: Concept of Kinetic Energy 

The energy stored in the body by virtue of its motion, is called kinetic energy.

The energy stored in a body by virtue of its position is called the gravitational potential energy of the system. The gravitational potential energy is calculated as-

where,

U_g = Gravitational Potential energy

g= Gravitational acceleration

R= Displacement in meter

m= Mass of body.

calculation

Part (a):

Let us take for the particle-bowl-Earth system when the particle is at B. Since and, at B, kinetic energy is given as-

KB=12mv2B  =12(0.200 kg) (1.50) =0.255J

Part(b):

At A, and the whole energy at A is:

At A, vi=0, KA=0and the whole energy at A is:

If no non-conservative forces act within the isolated system, the mechanical energy of the system is not conserved, so

At B

Part (c):

No, we cannot determine.

Part (d):

It is possible to find an effective coefficient of friction, but not the actual value of since normal reaction and frictional force both vary with position.