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(D) $1$. When the object slides down with uniform velocity, the net force is zero. Thus, the component of gravity $mg \sin 	heta$ is balanced by the

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stays at rest on the inclined plane and will not slide down.

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(D) $1$. When the object slides down with uniform velocity, the net force is zero. Thus, the component of gravity $mg \sin 	heta$ is balanced by the kinetic friction $f_k = \mu_k N = \mu_k mg \cos 	heta$. Therefore, $mg \sin 	heta = \mu_k mg \cos 	heta$, which implies $\mu_k = 	an 	heta$.$2$. When the object is pushed up with initial velocity $u$, it moves against both gravity and friction. The retardation is $a = g \sin 	heta + \mu_k g \cos 	heta = g \sin 	heta + (	an 	heta) g \cos 	heta = 2g \sin 	heta$.$3$. After stopping at the top, the object slides down. The acceleration while sliding down is $a' = g \sin 	heta - \mu_k g \cos 	heta = g \sin 	heta - (	an 	heta) g \cos 	heta = 0$. Wait, this implies the object would remain at rest if the static friction equals kinetic friction. However, since it was already sliding, the net force down the plane is $mg \sin 	heta - f_k = 0$. Thus, the object will remain at rest at the point where it stops.

When the angle of inclination of an inclined plane is $\theta$, an object slides down with uniform velocity. If the same object is pushed up with an initial velocity $u$ on the same inclined plane, it goes up the plane and stops at a certain distance on the plane. Thereafter, the body:

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