Question

An 160.0 g hockey puck slides along an essentially frictionless ice rink with speed 4.70 m/s. A hockey player uses her stick to do –1.20 J of work on the puck. What is the puck's speed after she has done this work?

A. 0 m/s

B. 2.66 m/s

C. 3.50 m/s

D. 4.53 m/s

Answer 1

Using work-energy theorem:

Work-done = Change in KE

W = dKE

W = KEf - KEi

W = 0.5*m*Vf^2 - 0.5*m*Vi^2

Given that, W = -1.20 J

Vf^2 = [0.5*160*10^-3*4.70^2 - 1.20]/(0.5*0.16)

Vf^2 = 7.09 m/sec

Vf = sqrt (7.09)

Vf = 2.66 m/sec

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