Question

A body is sliding down a rough inclined plane which makes angle 30° with the horizontal. The coefficient of the friction between the surfaces of contact of the body and the plane is
0.25. Calculate the acceleration.
O 1.65 m per second sq..
O 3.42 m per second sq..

O 2.78 m per second sq..
O 1.4 m per second sq..

Answer 1

Given :

Angle of inclination = 30°

Coefficient of friction = 25

To Find :

Acceleration of the block.

Solution :

From the FBD of block, we can say that only two forces act on the line of motion$.$

First one is mgsinθ and second one is μN (friction force).

Net force on the block will be

→ mgsinθ - μN

As per newton's second law of motion force is measured as the product of mass and acceleration.

→ ma = mgsinθ - μ(mgcosθ)

→ a = g(sinθ - μcosθ)

→ a = 9.8(sin30° - 0.25cos30°)

→ a = 9.8(1/2 - 0.25 × √3/2)

→ a = 9.8(0.5 - 0.216)

→ a = 9.8(0.284)

→ a = 2.78 m/s²

∴ (C) is the correct answer!

Answer 2

Answer:

c = 2.78 m per second square.