Question

A 20 kg sled is coasting with constant velocity at 5 m/s over perfectly smooth, level ice. It enters a rough stretch of ice 20 m long in which the force of friction is 8 N. With what speed does the sled emerge from the rough stretch

Answer 1

Answer:

vf = 3 m/s

Explanation:

• Applying the work-energy theorem, we know that the change in kinetic energy of the object while crossing the rough part of the ice, is equal to the work done by the friction force (as it is the only net force doing work on the sled), as follows:

       $\Delta K = \frac{1}{2}* m* (v_{f}^{2} -v_{o} ^{2} ) = W_{Ffr} (1)$

Replacing for the givens, we get:

$\Delta K = \frac{1}{2}* 20kg* (v_{f}^{2} -(5 m/s) ^{2} ) = -20 m * 8 N (2)$

• Rearranging terms, and solving for vf, we get:

$v_{f} = \sqrt{-16 (m/s)2 + 25 (m/s)2} = \sqrt{9 (m/s2)} = 3 m/s$

vf = 3 m/s