Question

1. A 1 kg hockey puck is moving at 10 m/s across the ice. At the other end of the ice rink, its speed is measured at 9.2 m/s. If the puck traveled 60

m, how much friction force acted on the puck? (assume the friction force was constant the whole time).

Answer 1

0.128 N

Step-by-step explanation:

½ mv²initial = ½ mv²final + Ff d

½ (1)(10²) = ½ (1)(9.2²) + (Ff)(60)

50 = 43.32 + (Ff)(60)

7.68 = (Ff)(60)

0.128 = Ff

Note:

Ff - force of friction