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Oxana [17]
3 years ago
12

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 inc

line, it's speed is vf. At what distance from the top of the incline is the speed of the block 0.5vf?
Physics
1 answer:
erica [24]3 years ago
8 0
Since 
<span>Vf^2 = 2*a*S </span>
<span>Given S=3.6m, thus </span>
<span>a = Vf^2/(2*3.6) </span>
<span>a = Vf^2/7.2 </span>

<span>Let d be the distance along the slope at which the velocity is 0.5Vf, then </span>
<span>(0.5Vf)^2 = 2*a*d </span>
<span>or </span>
<span>d = (0.5*Vf)^2/(2*a) </span>
<span>with a = Vf^2/7.2, we have </span>
<span>d = 0.9 m</span>
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