A mass that is 10 kg slides down an inclined plane. The coefficient of kinetic f

Question

A mass that is 10 kg slides down an inclined plane. The coefficient of kinetic friction is uk=0.120. At the bottom of the incline, the mass slides over a frictionless horizontal plane before colliding with a compressing spring of spring constant k=50000 N/m. The spring obeys Hooke's Law. The angle of the incline is 45 degrees. The length of the incline is 5 m and it is 5 m in height.

By using the work-energy theorem, calculate the kinetic energy and hence the velocity of the mass just as it touches the spring.

Explanation / Answer

initial potential energy = 5*10*10 = 500
frictional force = mgcos 45 *uk = 50*1/2*.12

distance travelled by block = 52

power dissipated = distance *F = 6/2*52 =30

final potential energy of spring = 500-30 = 470

1/2kx^2 = 470

x^2 = 940/50000

x = .137m = 137cm

for velocity before touching spring equate its kinetic energy

1/2mV^2 = 470

V^2 = 940/10

V = 9.69

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