A 50 kg block is released from rest and then slides for a distanceof 2.0 meters

Question

A 50 kg block is released from rest and then slides for a distanceof 2.0 meters down a ramp that is inclined 45 degrees above thehorizontal. The 50 kg block then strikes and ideal spring that isanchored to the bottom of the ramp. The block causes the spring to.Be compressed 15 cm before momentarily coming to rest. The surfaceof the ramp has a coefficient of kinetic friction of 0.1

A. What is the speed of the 50 kg block just prior to contactingthe ideal spring?

B. What is the spring constant of the ideal spring?

Please show final answer and all work done to get that answer

Explanation / Answer

To calculate the speed of the block just prior to contactingthe spring, you must first find the acceleration Fnet = F(parallel) - Ff Fnet = mgsin - mgcos      =346.48 - 34.648 = 309.832N 309.832/50kg = 6.1966m/s2 v2 = vo2 + 2a(x) v2 = 2(6.1966m/s)(2m) = v = 4.98m/s The force that the block will be hitting the spring with isthe Fnet that was already calculated. Fnet =-kx = 309.832/.15m = 2065.55N/m v2 = vo2 + 2a(x) v2 = 2(6.1966m/s)(2m) = v = 4.98m/s The force that the block will be hitting the spring with isthe Fnet that was already calculated. Fnet =-kx = 309.832/.15m = 2065.55N/m

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