Question #221753

A block is placed on top of a fixed smooth inclined plane inclined at 30° to horizontal. The length of plane is 5 m. The block slides down the plane and reaches at bottom. The speed of the block at bottom will be nearly


1
Expert's answer
2021-08-03T11:42:49-0400

By the condition of the problem, we have a smooth inclined plane.\text{By the condition of the problem, }\newline \text{we have a smooth inclined plane.}

That is, frictional forces can be neglected.\text{That is, frictional forces can be neglected.}

l=5ml = 5m

h=lsin30°=2.5mh = l*\sin30\degree= 2.5m

E=Ep+EkE = E_p+E_k

At the top of an inclined plane:\text{At the top of an inclined plane:}

Ep=mgh;Ek=0E_p = mgh;E_k=0

At the foot of an inclined plane:\text{At the foot of an inclined plane:}

E=Ep+EkE = E_p'+E_k'

Ek=mv22;Ep=0;E_k'=\frac{mv^2}{2};E_p'=0;


mgh=mv22mgh = \frac{mv^2}{2}

v=2gh=29.82.5=7msv=\sqrt{2gh}=\sqrt{2*9.8*2.5}=7\frac{m}{s}


Answer: v=7ms\text{Answer: }v=7\frac{m}{s}


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