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
(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,
The angle of inclination,
θ = 15.0 °
Therefore acceleration of the block is:
We can find the speed of the block when it reaches the bottom of the incline by using kinematic equation,
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,
Length of the inclined surface,
s=2.00 m
Acceleration of the block,
Substituting the values in above equation we can find the speed of the block when it reaches the bottom of incline.
Comments
Leave a comment