Question

A block of mass 1.59 kg is connected to a spring of spring constant 148 N/m which is then set into oscillation on a surface with a small coefficient of kinetic friction. The mass is pulled back 30.6 cm to the right and released. On the first right to left oscillation, the mass reaches 29.38 cm to the left.

Part A

What is the coefficient of friction?

Part B

To what distance does the mass return on the slide back to the right?

Part C

What is the total energy lost by the system in the first full oscillation?

Give your answer in Joules.

Part D

What percentage of the total energy was lost in the first full oscillation?

Answer 1

Part A) The Amplitude of SHM is 30.6 cm = 0.306 m

Hence its maximum potential energy = (1/2)kA^2 = (1/2)*148*0.306^2 = 6.929 J

Its potential energy at 0.2938 m is,

(1/2)*148*0.2938^2 = 6.387 J

Hence work done by friction is 6.929 - 6.387 = 0.542J

Work done by friction = Frictional force*distance

0.542 = μmg*(0.306+0.2938)

0.542 = μ*1.59*9.81*0.5998

=> μ = 0.058

Part B) Energy at the starting point is 6.387J(as calculated above)

Frictional force = μmg = 0.058*1.59*9.81 = 0.9046 N

Let the particle travel a distance x to the right.

Then its potential energy at x is (1/2)*148x^2 = 74x^2

Work done by friction = 6.387 - 74x^2

0.9046*(0.2938+x) = 6.387 - 74x^2

0.2658 + 0.9046x = 6.387 - 74x^2

74x^2 + 0.9046x - 6.1212 = 0

Therefore x = 0.2815 m = 28.15 cm

Part C: Potential energy at 0.2815 m is

(1/2)*148*0.2815^2 = 5.86J

Hence energy lost is initial energy - final energy = 6.929 - 5.864 = 1.065 J

Part D) 100* 1.065/6.929 = 15.37 %