A hockey puck with mass 0.16 kg is at rest at the origin (x = 0) on the horizont

Question

A hockey puck with mass 0.16 kg is at rest at the origin (x = 0) on the horizontal, frictionless surface of an ice rink. At time t = 0, a player applies a force of 0.280 N to the puck, parallel to the x-axis; he continues to apply this force until t = 2.10 s.

What is the speed of the puck in m/s at t = 2.10 s?

What is the position of the puck in meters at t = 2.10 s?

If the force ceases after t = 2.10 s and then restarts at 5.10 s, what is the speed of the puck in m/s at 6.50 s?

What is the position of the puck in meters at t = 6.50 s?

Explanation / Answer

F = ma a = F/m a = 0.280/0.16 a = 1.75 m/s/s x = 1/2at^2 x = 1/2(1.75)(2.10)^2 x = 3.858 m between 2.10 and 5 seconds, the speed stays at 3.675 m/s and the distance v = at v = 1.75(2.10) v = 3.675 speed of the puck in m/s at t = 2.10 s is 3.675 m/s position of the puck in meters at t = 2.10 s is 3.858 m in the 3 seconds between time t = 2.10 and t = 5.10 the puck travels d = 3.675(3) = 11.025 m distance for the last two seconds is d = v0t + 1/2at^2 d = 3.675(2.10) + 1/2(1.75)(2.10^2) d =11.576 so the total distance from the origin is d = 3.858 + 11.576 + 11.025 d = 26.459 m There was a total of 3.5 seconds of acceleration v = at v = 1.75(4) v = 6.125 m/s

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