Ignoring ball rotation and friction, the work done is F * X and equals the final kinetic energy, (1/2) MV^2
Therefore F = (1/2)MV^2/X
X= 22m
When rotational kinetic energy is considered, (3/5)MV^2 replaces (1/2)MV^2, but they probably do not expect you to know that.
A steel ball has a mass of 4.0kg and rolls along a smooth, level surface at 62m/s.
A) Find its kinetic energy(i got this and its 7688 joules)
B)?????-At first, the ball was at rest on the surface. A constant force acted on it through a distance of 22 meters to give it the speed of 62m/s. What was the magnitude of the force??
I really don't understand part b. please help in detail! Thxs PS- i need to use energy work formuals. no kinematics formuals
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so whats the final answer??
I gave you the formula for F. Finish it yourself
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