A mass of 7 kg is compressed into a spring on a horizontal surface. It is compre

Question

A mass of 7 kg is compressed into a spring on a horizontal surface. It is compressed into the spring 0.55 meters and held still. The force required to do this is 644 Newtons. It is then released from rest and the mass travels over a originally frictionless, horizontal surface, but at the end of the surface there is some friction with a coefficient of kinetic friction of 0.74. It then travels up a frictionless incline with an inclination angle of 16 degrees. It travels a distance of 3 meters up the incline before it stops. What was the length, in meters, of the portion of the horizontal surface that had friction?

Explanation / Answer

This problem can be solved using the principle of conservation of energy.

First determine the spring constant K of the spring.

F = K*x

=> 644 = K*0.55

=> K = 644 / 0.55 = 1171 N/m

Now, the initial potential potential energy of the spring, Ep = (1/2)*K*x^2 = 0.5*1171*0.55^2 J

Again, this will be equal to work done against friction and potential energy attained by the mass.

This is equal to, Ew = u*mg*d + mg*3*sin16

equalize the two -

0.5*1171*0.55^2 = u*mg*d + mg*3*sin16

=> 177.1 = 0.74*7*9.8*d + 7*9.8*3*sin16 = 50.76*d + 56.73

=> d = (177.1 - 56.73) / 50.76 = 2.37 m

So, the length of the horizontal surface that has friction = 2.37 m

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