A block of mass 1.4 kg is attached to a horizontal spring that has a force const

Question

A block of mass 1.4 kg is attached to a horizontal spring that has a force constant 1 200 N/m as shown in the figure below. The spring is compressed 2.0 cm and is then released from rest. (a) A constant friction force of 3.2 N retards the block's motion from the moment it is released. Using an energy approach, find the position x of the block at which its speed is a maximum. cm (b) Explore the effect of an increased friction force of 14.0 N. At what position of the block does its maximum speed occur in this situation? The correct answer is not zero.

Explanation / Answer

Let's say the origin is at the tip of the spring when it is completely relaxed. Then its potential energy of compression is kx^2/2. At any point of decompression, the reduction in potential energy is converted to the K.E of the block, after being lessened by the frictional energy loss. So then: 0.5k(0.02^2 - x^2) - F(0.02 - x) = 0.5mv^2 v = sqrt ( ( k(0.02^2 - x^2) - 2F(0.02 - x) ) / m ) dv/dt = 0.5 (2kx - 2F) / sqrt ( ( k(0.02^2 - x^2) - 2F(0.02 - x) ) / m ) for maximum velocity, dv/dt = 0, so 2kx - 2F = 0 x = F/k

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