STATEMENT-1: A block of mass m starts moving | vedclass

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(C) In the first case (horizontal surface),the work done by friction is $W_1 = -f_k d_1 = -\mu mg d_1$. The block stops when its initial kinetic energ

Vedclass · Accepted Answer

$Statement-1$ is True,$Statement-2$ is False.

Vedclass · Answer

(C) In the first case (horizontal surface),the work done by friction is $W_1 = -f_k d_1 = -\mu mg d_1$. The block stops when its initial kinetic energy is dissipated: $\frac{1}{2}mv^2 = \mu mg d_1$,so $d_1 = \frac{v^2}{2\mu g}$. The decrease in mechanical energy is $\Delta E_1 = \frac{1}{2}mv^2$.In the second case (inclined surface),the work done by friction is $W_2 = -f_k d_2 = -\mu mg \cos(30^{\circ}) d_2$. The block stops when its initial mechanical energy is dissipated: $\frac{1}{2}mv^2 = \mu mg \cos(30^{\circ}) d_2 + mg d_2 \sin(30^{\circ})$. The decrease in mechanical energy is the work done by friction,$\Delta E_2 = \mu mg \cos(30^{\circ}) d_2$. Since $\cos(30^{\circ}) < 1$,the frictional force is smaller,and the distance $d_2$ is also different. However,the coefficient of friction $\mu$ is a property of the materials and does not change with the angle of inclination. Thus,$Statement-2$ is False. Since the energy dissipated by friction is $\Delta E = \int f_k dx$,and $f_k = \mu N$,where $N = mg \cos(	heta)$,the energy loss is indeed smaller in the second case. Thus,$Statement-1$ is True.

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