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Question

A block slides down a frictionless plane having an inclination of $$\theta= 15.0^{0}$$. The block starts from rest at the top, and the length of the incline is $$2.00 m$$. (a) Draw a free-body diagram of the block. Find (b) the acceleration of the block and (c) its speed when it reaches the bottom of the incline.

Solution
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(a) Below figure shows the forces on the object. The two forces acting on the block are the normal force, $$n$$, and the weight, $$mg$$. If the block is considered to be a point mass and the $$x$$ axis is chosen to be parallel to the plane, then the free-body diagram will be as shown in the figure below. The angle $$\theta$$ is the angle of inclination of the plane.
Applying Newton’s second law for the accelerating system (and taking the direction up the plane as the positive $$x$$ direction), we have
$$\sum F_{y} = n − mg \cos\theta = 0 : n = mg \cos\theta$$
$$\sum F_{x} = −mg \sin\theta = ma : a = –g \sin\theta$$
(b) When $$\theta = 15.0^{0}$$,
$$a = −2.54 m/s^{2}$$.
(c) Starting from rest,
$$v_{f}^{2}=v_{i}^{2}+2a(x_{f}-x_{i})=2a\Delta x$$
$$|v_{f}|=\sqrt{2|a|\Delta x}=\sqrt{2|-2.54m/s^{2}|(2.00m)}=3.19m/s$$

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