A block with mass m = 13 kg rests on a frictionless table and is accelerated by

Question

A block with mass m = 13 kg rests on a frictionless table and is accelerated by a spring with spring constant k = 4971 N/m after being compressed a distance x1 = 0.111 m from the spring’s unstretched length. The floor is frictionless except for a rough patch a distance d = 2.6 m long. For this rough path, the coefficient of friction is µk = 0.5.

~How far will the block slide into the rough path before coming to rest?

I know the work that the spring does, the force of the spring, the force of friction of the path, and the work that the path is capable of doing. I know I am looking for a distance, and have tried several things: Using W=f*d for the net force and work being applied, and using the kinematics equation v^2(final)=v^2(initial)=2ax [v^2(initial) being 0, and v^2 final using KE=1/2mv^2, work being the change in kinetic energy] - however, I am arriving at incorrect answers after many attempts. Thanks in advance.

Explanation / Answer

0.5*k*x^2 = 0.5*m*v^2 = µmgs Therefore s = (0.5*k*x^2)/(µmg) = 0.48 m

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