Question

A moving 3.70kg block collides with a horizontal spring whose spring constant is 363 N/m. The block compresses the spring a maximum distance of 13.50cm from its rest postion. The coefficient of kinetic friction between the block and the horizontal surface is 0.310. a) What is the work done by the spring in bringing the block to rest? Hint: "be careful about the sign of the work! Is it positive or negative?" b) How much mechanical energy is being dissipated by the force of friction while the block is being brought to rest by the spring? c) What is the speed of the block when it hits the spring?

Answer 1

Part A.

Using work-energy theorem:

Work-done by spring = Change in spring energy

Ws = 0.5*k*xf^2 - 0.5*k*xi^2

xi = 0 m and xf = 13.50 cm =0.135 m

k = spring constat = 363 N/m

So,

Ws = 0.5*363*0.135^2 - 0.5*363*0^2

Ws = 3.31 J

Part B.

Work-done by frictional force will be

Wf = Ff*d

Wf = uk*N*d

Wf = uk*m*g*d

Wf = 0.310*3.70*9.81*0.135

Wf = 1.52 J

Part C.

Again Using work-energy theorem:

Total Work = Change in kinetic energy

W = Ws + Wf = 3.31 + 1.52

dKE = KEf - KEi

KEi = 0, since initially at rest

W = KEf

3.31 + 1.52 = 0.5*m*Vf^2

Vf = sqrt [2*(3.31 + 1.52)/3.70]

Vf = 1.62 m/sec

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