Question

A block is lying on an inclined plane which makes 60 degree

with the horizontal. If coefficient of friction between

block and plane is 0.25 and g = 10m/s2

, then

acceleartion of the block when it moves along the

plane will be

Answer 1

Acceleration of the block will be 7.41 m/s^2

Let the box moves along the plane with the acceleration a

The weight of the block along the inclined plane is given as

W=mgsinθ

The friction force which oppose the movement of the block is given as

F=μmgcosθ

Therefore the net force on the block is

F_net=ma=( mgsinθ-μmgcosθ)

a=gsinθ-μgcosθ

Plugging the values in the above equation

a=10sin60-0.25*10cos60

=7.41 m/s^2

Therfore the acceleration of the block is 7.41 m/s^2

Answer 2

Let the box moves along the plane with the acceleration a

The weight of the block along the inclined plane is specified as W=mgsinθ

The friction force which contest the movement of the block is given as F=μmgcosθ

Therefore, the net force on the block is F.net=ma= (mgsinθ-μmgcosθ)

a=gsinθ-μgcosθ

Putting the values in the equation, we get

A = 10sin60-0.25*10cos60 =7.41 m/s^2

Therefore, the acceleration of the block is 7.41 m/s^2.