# A 100 N box falls from a truck moving at 25 m/s and slides to a stop in 55 m. What is the coefficient of friction between the box and the highway. How much work was done on the box by the friction?

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A 100 N box falls from a truck moving at 25 m/s. The kinetic energy of the box when it falls is (1/2)*(100/9.8)*25^2 = 3.188 kJ. A frictional force acts on the box when it lands on the highway that is opposite in direction to the direction of of movement of the box. The magnitude of the force is Fc = C*N where C is the coefficient of friction and N is the normal force. The work done on the box by the force of friction as it slides 55 m is C*100*55. This has to be equal to the initial kinetic energy of the box to bring it to a stop. The work done on the box is approximately 3.188 kJ

Equating the initial kinetic energy and the work done by the force of friction gives:

C*100*55 = (1/2)*(100/9.8)*25^2

=> C = 0.579

The workdone on the box by the force of friction

**The coefficient of friction between the box and the highway is 0.579**