Question

A block slides down a frictionless plane having an inclination of 0 = 15.0°. 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 blockand (c) its speed when it reaches the bottom of the incline

Answer 1

(a) The figure represents the free-body diagram of the block that slides on frictionless inclined surface.

(b) The weight of the block is resolved into two components. The component

mg cos θ

is equal to the normal force and the component

mg sin θ

is the force that slides the block on the inclined surface. Here

θ

is angle of inclination. From Newton’s law the net force is equal to the product of mass

(m)

and acceleration

(a)

Therefore,

$F_{net} = mg sin θ \\ ma=mg sin θ \\ a = g sin θ$

The angle of inclination,

θ = 15.0 °

Therefore acceleration of the block is:

$= 9.80 \times sin 15.0° \\ = 2.54 \;m/s^2$

We can find the speed of the block when it reaches the bottom of the incline by using kinematic equation,

$v_f^2 -v_i^2 = 2as$

v_f is the final speed of the block at the bottom of incline

s is the length of the incline

The block starts from rest at the top of inclined surface. So, the initial speed of the block at the top of inclined surface,

$v_i=0$

Length of the inclined surface,

s=2.00 m

Acceleration of the block,

$a=2.54 \; m/s^2$

Substituting the values in above equation we can find the speed of the block when it reaches the bottom of incline.

$v_f^2 -v_i^2 = 2as \\ v_f^2 -0 = 2 (2.54)(2.00) \\ v_f = \sqrt{2 \times 2.54 \times 2.00} \\ = 3.18 \;m/s$