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CaptainADHD
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Homework Statement
Starting from rest at t = 0, a 5.0-kg block is pulled across a horizontal surface by a constant horizontal force having a magnitude of 12 N. If the coefficient of friction between the block and the surface is 0.20, at what rate is the 12-N force doing work at t = 5.0 s?
Homework Equations
Force = mass x acceleration
Friction force = coeff of friction x normal force
Work = mass x acceleration x distance
Power = mass x acceleration x distance / time ... ?
The Attempt at a Solution
Force in the horizontal direction should be applied force minus friction force right? So:
12.0 Newtons - (0.20 x 5.0 kg x 9.8) Newtons = 2.20 Newtons
acceleration in x direction = 2.2 Newtons / 5 kg = .44 meters per second squared
final velocity = acceleration x time = .44 x 5 = 2.2 m/s
power = force x speed = 2.2 Newtons x 2.2 meters per second = 4.84 ?
or... mass x acceleration x distance = work per second? That would be:
5 kg x .44 m/s^2 x 5.5 m / 1 s = 12.1 watts ?
0.13 kW
0.14 kW
0.12 kW
26 W
12 W
What I don't understand is what "power at time x" means. Wouldn't that be instantaneous power? I thought power was total work over a time interval. I'm confused as hell ><