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Energy Conservation for Spring on Rough Surface

Unit: Work, Energy, and Power

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Part 1

Consider the block–spring–ramp system shown below.

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A block of mass mm is attached to a horizontal spring of force constant kk fixed to a wall on a horizontal table. The block is initially pushed toward the wall, compressing the spring by distance x0x_0, and then released from rest. While the block is in contact with the spring, friction is negligible.

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Immediately to the right of the spring region is a horizontal surface of length LL with coefficient of kinetic friction μk\mu_k. Beyond this rough section, the block encounters a frictionless ramp that makes angle θ\theta with the horizontal. The acceleration due to gravity is gg.

The parameter values are

m=1.22kg, k=184N/m, x0=0.28m,L=1.87m, μk=0.21, θ=28, g=9.8m/s2.m = 1.22 \,\mathrm{kg},\ k = 184 \,\mathrm{N/m},\ x_0 = 0.28 \,\mathrm{m}, \\ L = 1.87 \,\mathrm{m},\ \mu_k = 0.21,\ \theta = 28^{\circ},\ g = 9.8 \,\mathrm{m/s^2}.

Using conservation of energy between the instant the block is at rest with the spring compressed by x0x_0 and the instant the spring has just returned to its natural length and lost contact with the block, determine the speed v1v_1 of the block immediately after it leaves the spring.

Correct!

Solution:

With friction negligible while the block is in contact with the spring, mechanical energy is conserved between the initial compressed position and the instant the spring returns to its natural length.

Initially, the block is at rest and the energy is purely spring potential:

Ki=0,Us,i=12kx02.K_i = 0, \qquad U_{s,i} = \tfrac12 k x_0^2.

When the spring reaches its natural length and loses contact, the spring potential is zero and the energy is purely kinetic:

Kf=12mv12,Us,f=0.K_f = \tfrac12 m v_1^2, \qquad U_{s,f} = 0.

Conservation of mechanical energy gives

Ki+Us,i=Kf+Us,fK_i + U_{s,i} = K_f + U_{s,f} 0+12kx02=12mv12+0.0 + \tfrac12 k x_0^2 = \tfrac12 m v_1^2 + 0.

Solving for v1v_1:

v1=x0km.v_1 = x_0 \sqrt{\frac{k}{m}}.

Thus,

v1=3.44m/s.v_1 = 3.44\,\mathrm{m/s}.

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