please show steps thank you A 2.40-kg block is released down a 25degree incline

Question

please show steps thank you

A 2.40-kg block is released down a 25degree incline with a friction coefficient of mu = 0.30 onto a spring from a vertical height of h = 5.00 m. When the block is momentarily at rest, the spring is compressed by 3.00 m. Find the speed of the block when the compression of the spring is 1.50 m. Follow-Up Challenge Question: The block is replaced with a 3.00 kg block, using the same incline, the same spring, and the same mu What is the most the spring gets compressed now?

Explanation / Answer

Here, 1/2 *2.40 * v2 = 2.40 * 9.8 * 5 - 0.30 * 2.40 * 9.8 * cos25 * (5/sin25)

=>   v = 5.91 m/sec

=>    1/2 * k * 3 * 3   + 0.30 * 2.40 * 9.8 * cos25 * 3 = 1/2 *2.40 * 5.912

=> spring constant , k = 5.051 N/m

=>   1/2 * 5.051 * 1.5 * 1.5 + 0.30 * 2.40 * 9.8 * cos25 * 1.5 = 1/2 *2.40 * (5.912 - v2)

=>   v   = 4.711 m/sec

=> speed of the block =   4.711 m/sec

Here,   1/2 * 5.051 * x2 + 0.30 * 3 * 9.8 * cos25 * x = 1/2 *3 * 5.912

=> most spring get compressed , x = 3.238   m

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