A block of mass 5 text{ kg} starts up a 45^{c | vedclass

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(A) The block stops on the inclined plane when its initial kinetic energy is used to do work against friction and is partly converted into potential e

Vedclass · Accepted Answer

$\frac{4 \sqrt{2}}{3} 	ext{ m}$

Vedclass · Answer

(A) The block stops on the inclined plane when its initial kinetic energy is used to do work against friction and is partly converted into potential energy.Given the angle of inclination is $	heta = 45^{\circ}$,the distance covered on the incline $(s)$ is related to the vertical height $(h)$ as:$\sin 45^{\circ} = \frac{h}{s} \implies s = \frac{h}{\sin 45^{\circ}} = h \sqrt{2}$.According to the work-energy theorem,the initial kinetic energy $(K.E.)$ is equal to the work done against friction plus the gain in potential energy:$K.E. = W_{	ext{friction}} + \Delta U$$K.E. = (\mu mg \cos 	heta) \cdot s + mgh$Substituting $s = h \sqrt{2}$ and $\cos 45^{\circ} = \frac{1}{\sqrt{2}}$:$100 = (0.5 	imes 5 	imes 10 	imes \frac{1}{\sqrt{2}}) 	imes (h \sqrt{2}) + (5 	imes 10 	imes h)$$100 = (0.5 	imes 5 	imes 10 	imes h) + 50h$$100 = 25h + 50h$$100 = 75h$$h = \frac{100}{75} = \frac{4}{3} 	ext{ m}$.Now,the distance covered $s$ is:$s = h \sqrt{2} = \frac{4}{3} \sqrt{2} 	ext{ m}$.

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