A 2.0{\ m kg} box moves back and forth on a horizontal frictionless surface betw

Question

A 2.0{ m kg} box moves back and forth on a horizontal frictionless surface between two different springs, as shown in the accompanying figure. The box is initially pressed against the stronger spring, compressing it 3.8{ m cm} , and then is released from rest.(Figure 1)

FIGURE 1: http://session.masteringphysics.com/problemAsset/1263189/4/1027560.jpg

Part A
By how much will the box compress the weaker spring?
Express your answer using two significant figures.
Delta l = ___________{ m cm}

Part B
What is the maximum speed the box will reach?
Express your answer using two significant figures.
v_{ m max} = ___________{ m m/s}

Explanation / Answer

Does the "accompanying figure" show the spring constants of both springs? The potential energy at the start is (1/2)*k1*4^2. This is all converted to kinetic energy while the box is between springs KE = (1/2)*m*v^2 = (1/2)*k1*4^2 solve for v This kinetic energy is converted back into potential energy in the second spring PE = (1/2)k2*(X2)^2 = (1/2)*k1*4^2 The compression of the second spring is: X2 = sqrt(16*k1/k2) You've got to know the k's... Also, I'm assuming that the box is not in contact with both springs at once, that should be shown in the figure. If that's not the case, the equation given for the Kinetic Energy is wrong. The assumption is that all of the Potential Energy (PE) of the first spring is transferred to the block, in the form of Kinetic Energy, and then the block contacts the second spring, transferring its kinetic energy back into PE in the second spring. If the box contacts the weaker spring before losing contact with the stronger spring, some of the PE of the first spring is transferred directly to the weaker spring, limiting the KE acquired by the block.

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