Question

A 1360.0 g mass is on a horizontal surface with μ_k = 0.33, and is in contact with a massless spring with a force constant of 781.0 N/m which is compressed. When the spring is released, it does 12.65 J of work on the mass while returning to its equilibrium position. Calculate the distance the spring was compressed.

What is the velocity of the mass as it loses contact with the spring?

Answer 1

1) Calculate the distance the spring was compressed.

Spring energy = 12.65 = (1/2)Kx^2

where

K = spring constant = 781 N/m (given)

x = distance at which spring was compressed

Therefore,

12.65 = (1/2)(781)(x^2)

Solving for "x"

x = 0.18 m

2) What is the velocity of the mass as it breaks contact with the spring?

Energy from spring = Energy absorbed by block + Energy to overcome friction

Es = Eb + Ef

12.65 = (1/2)mV^2 + µmgx

where

m = mass of the block = 1.360 kg. (given)

V = velocity at which block leaves the spring

µ = coefficient of friction = 0.33 (given)

g = acceleration due to gravity = 9.8 m/sec^2 (constant)

x = 0.18 m (as calculated above)

Substituting appropriate values,

12.65 = (1/2)(1.360)V^2 + (0.33)(1.360)(9.8)(0.18)

Solving for "V"

V = 4.18 m/s