A 6.0 kg block, starting from rest, slides down a frictionless incline of length 2.0 m. When it arrives at the bottom of the incline, it's speed is vf. At what distance from the top of the incline is the speed of the block 0.5vf?

Question

Mason Goodman

Physics

9 June, 10:39

A 6.0 kg block, starting from rest, slides down a frictionless incline of length 2.0 m. When it arrives at the bottom of the incline, it's speed is vf. At what distance from the top of the incline is the speed of the block 0.5vf?

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Answers (1)

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Solomon Avila · Answer 1

Since

Vf^2 = 2*a*S

Given S=3.6m, thus

a = Vf^2 / (2*3.6)

a = Vf^2/7.2

Let d be the distance along the slope at which the velocity is 0.5Vf, then

(0.5Vf) ^2 = 2*a*d

or

d = (0.5*Vf) ^2 / (2*a)

with a = Vf^2/7.2, we have

d = 0.9 m