A 2.5 kg block is used to compress a spring witha spring constant of 125 N/m by

Question

A 2.5 kg block is used to compress a spring witha spring constant of 125 N/m by 48cm. The spring and blockk are on a ramp inclined an angle 36.87 degrees above the horizontal.

a) If the ramp is smooth, how much farther up along the ramp will the block climb if it is unattached from the spring versus being affixed to it?

b) If the ramp is smooth, what coefficent of friciton is required to bring the block to rest once the spring has reached its equilibrium length?

ng constant of 125 N/m by 48.0 c with a spring horizontal. (a) If t an angle 36.87o the unattach the ramp will the block climb if it is b) If the ramp is not smooth, what coefficient once the spring has reached its equilibrium lengt

Explanation / Answer

The potential energy of the spring gets converted to gravitational PE of the block.

h = d sin(theta)

1/2 k x^2 = m g d sin(theta)

d = 0.5 k x^2/ m g sin(theta)

d = 0.5 x 125 x 0.48^2/2.5 x 9.8 x sin36.87 = 0.98 m

Hence the block climbs, d = 0.98 m

b)Thw rok done by the friction is:

W = u m g cos(theta) d

this should be equal to PE of the spring

u m g cos(theta)d = 0.5 k x^2

u = 0.5 k x^2/m g cos(theta) d

u = 0.5 x 125 x 0.48^2/ 2.5 x 9.8 x cos36.87 x 0.98 = 0.75

Hence, u = 0.75

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