A 2.00 kg block is attached to a spring, whose constant of the spring is 100 N/m

Question

A 2.00 kg block is attached to a spring, whose constant of the spring is 100 N/m. At a moment, it is observed that the spring is compressed by 0.100 m from its original position, and the speed of the block is 2.00 m/s pointing to the right at that moment. a) If the velocity of the block at current location is pointing to the right, what is the maximum distance the block can move rightward away from the current position when the surface is frictionless? b) If the velocity of the block at current location is pointing to the left, does the maximum distance calculated in part A change? c) How does the maximum distance change when the surface between the block and ground is frictional where the coefficient of kinetic friction is 0.100?

Explanation / Answer

a. if xm= maximum distance, then

1/2 k xm^2 = 1/2 kx^2 + 1/2mv^2

so, 1/2 x 100 x xm^2 = 1/2 x 100 x 0.1^2 + 1/2 x 2 x 2^2

so, xm= 0.3 m

b. yes maximum distance will change

c. in case of kinetic friction

1/2 x 100 x xm^2= 1/2 x 100 x 0.1^2+ 1/2x2x2^2 - 0.1 x 2x9.8 xm

so, 50 xm^2+1.96xm-4.5 =0

so, xm= 0.28 m

so maximum distance decreases by 0.3-0.28=0.02 m

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