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A small ball of mass m starts at a point A with speed $\;v_{0}\;$ and moves along a frictionless track AB as shown . The track BC has coefficient of friction $\;\mu\;$ . The ball comes to stop at C at travelling a distance L which is :

$(a)\;\large\frac{2h}{\mu}+\large\frac{v_{0}^{2}}{2 \mu g} \qquad(b)\;\large\frac{h}{\mu}+\large\frac{v_{0}^{2}}{2 \mu g} \qquad(c)\;\large\frac{h}{2\mu}+\large\frac{v_{0}^{2}}{ \mu g} \qquad(d)\;\large\frac{h}{2\mu}+\large\frac{v_{0}^{2}}{2 \mu g} $

1 Answer

Applying conservation of energy, we get mgh + 1/2mv²= (mew)mgL Solving we get L= h/(mew) + v²/2(mew)g
answered May 25 by mecool.dg
 

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