Question

6. The 200 g particle (Fig. 4) is released from point A at rest. The radus is 30 cm. The speed of the particle at point B is 1.5 m/s. (a) What is its kinetic energy at B? (b) How much energy is lost as a result of friction as the particle goes from A to B? (c) Is it possible to determine μ from these results in any simple manner? Explain. Figure 4

Answer 1

a)
Kinetic energy = 1/2 mv²
Where v is the velocity at B
= 0.5 x 0.2 x (1.5)²
= 0.225 J

b)
Using conservation of energy, energy at B = energy at A
E_B = mgR
= 0.2 x 9.81 x 0.3
= 0.587 J

Energy lost = E_B - original kinetic energy at B
= 0.587 - 0.255
= 0.334 J

c)
Frictional force at A is zero and the frictional force at B is _$\mu$mg
Average frictional force, Ff = 1/2 _$\mu$mg
Work done by frictional force = Ff x ($\pi$/2) R
Where ($\pi$/2)R is the distance travelled from A to B.
Work done = 1/2 _$\mu$mg x ($\pi$/2) R
= 0.5 x _$\mu$ x 0.2 x 9.81 x 1.571 x 0.3
= 0.462 _$\mu$

Energy lost = Work done by frictional force
0.334 = 0.462 _$\mu$
$\mu$ = 0.334/0.462
= 0.787