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(B) $1$. When the object slides down with uniform velocity,the net force is zero. This implies the component of gravity $mg \sin 	heta$ is balanced b

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slides down the inclined plane and reaches the ground with velocity less than $u$.

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(B) $1$. When the object slides down with uniform velocity,the net force is zero. This implies the component of gravity $mg \sin 	heta$ is balanced by the kinetic friction $f_k = \mu_k N = \mu_k mg \cos 	heta$. Thus,$\mu_k = 	an 	heta$.$2$. When the object is pushed up with velocity $u$,it experiences a retardation $a_{up} = g \sin 	heta + \mu_k g \cos 	heta$. Since $\mu_k = 	an 	heta$,$a_{up} = g \sin 	heta + (	an 	heta) g \cos 	heta = 2g \sin 	heta$.$3$. After stopping,the object slides down. The acceleration while sliding down is $a_{down} = g \sin 	heta - \mu_k g \cos 	heta$. Since $\mu_k = 	an 	heta$,$a_{down} = g \sin 	heta - g \sin 	heta = 0$. Wait,this implies uniform velocity. Let's re-evaluate: the kinetic friction acts opposite to motion. When sliding down,$a_{down} = g \sin 	heta - \mu_k g \cos 	heta = 0$. The object will slide down with the same velocity it had at the start of the descent. Since it started from rest at the top,it will slide down with uniform velocity. However,the question implies it reaches the ground. The energy lost due to friction during the upward journey is $W_f = f_k 	imes d = (\mu_k mg \cos 	heta) d$. During the downward journey,the same energy is lost. By work-energy theorem,the final kinetic energy $K_f = K_i + W_{gravity} - W_{friction}$. Since $W_{gravity}$ is the same for both paths and $W_{friction}$ is the same,the final velocity will be less than $u$ due to energy dissipation.

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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