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.
(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,
$$v_{f}^{2}=v_{i}^{2}+2a(x_{f}-x_{i})=2a\Delta x$$
$$|v_{f}|=\sqrt{2|a|\Delta x}=\sqrt{2|-2.54m/s^{2}|(2.00m)}=3.19m/s$$