# Numerical problems based on the inclined plane physics – solved

Last updated on April 15th, 2021 at 01:50 pm

In this post, we will focus on **numerical problems based on inclined plane physics**. We will list down the problems and solve them one by one.

## Numerical problems based on the inclined plane physics

1] A 5-kg mass, initially at rest, slides down a frictionless 30° incline. Calculate the force parallel to the incline and the acceleration of the mass. If the incline is 0.8 m long, calculate the velocity of the mass when it reaches the bottom of the incline.

1] solution:

The goal of the problem is to calculate the desired quantities of force, acceleration, and velocity parallel to the incline **(when friction is neglected).****The force parallel to the incline** =F= mg sin θ = (5) (9.8) sin 30 = 24.5 N downwards.**Acceleration of the mass** =a= g sin θ = (9.8) sin 30 = 4.9 m/s^{2} downwards

say the velocity of the mass when it reaches the bottom of the incline V and we have to find it out. Then we can use the formula like this:

V^{2} = U^{2} + 2 a S

(Here, *U =0, acceleration a = 4.9 m/s ^{2} and S = 0.8 m*)

So,

**V = (2 a S)**= (2 x 4.9 x 0.8)

^{1/2}^{1/2}= 2.8 m/s

2 ] A 2-kg mass is sliding down an incline with an unknown angle, as shown. The incline is assumed to be frictionless and is 1.2 m long. The mass starts from rest and is observed to take 2.3 s to reach the bottom. What is the angle of the incline?

Acceleration of the mass along the incline a = g sin θ = 9.8 sin θ …… (1)

We need to use this formula: **S = Ut + (1/2) a t ^{2}**

As, U = 0, the formula becomes:

**S = (1/2) a t**

^{2}=>

**a = (2 S)/ t**= (2 x 1.2) /2.3

^{2}**= 0.45 m/s**

^{2}**…. (2)**

^{2}Now from equation 1 and 2:

0.45 = 9.8 sin θ

=> sin θ = 0.45/9.8 = 0.046

=>

**θ = 2.63 degrees**