A cart is placed at the summit O of an inclined plane OB with length L = OB = 250 m and height h = OH = 50 m above the horizontal. The cart has a mass m = 100 kg. The frictional force f parallel to the plane is f = 20 N. The cart is initially pushed down the plane with velocity v0 = 1 m/s. Take g =9.8m/s2. What is the distance covered in terms of t

According to Newton's second law of motion, the net force (∑F) on the crate is equal to its mass (m) times its acceleration (a). After finding the acceleration, we can use kinematic equations, also known as SUVAT equations, to model the crate's position as a function of time. Drawing a free-body diagram will allow us to show all the forces acting on the crate.

According to the free-body diagram, there are three forces on the crate:

Normal force N pushing perpendicular to the inclined plane
Friction force f pushing parallel to the inclined plane
Weight force mg pulling down

Sum of forces in the parallel direction:

∑F = ma

mg sin θ − f = ma

a = g sin θ − f/m

a = g (h/L) − f/m

Plug in values and solve.

a = 9.8 m/s² (50 m / 250 m) − (20 N / 100 kg)

a = 1.76 m/s²

Next, use the kinematic equation:

s = ut + ½ at²

where

s is the displacement
u is the initial velocity
a is the acceleration
t is the time

Given u = 1 m/s and a = 1.76 m/s², the distance as a function of time is:

s = (1) t + ½ (1.76) t²

s = t + 0.88 t²