Question

A body slides down from rest from the top of a 6.4 m long inclined plane inclined at 30 degree . Find the time taken

by the body to reach the bottom of the plane. Take uk = 0.2 and g = 9.8 m s-2​

Answer 1

Solution:-

• Distance traveled by the body = 6.4 m

• Angle of inclination = 30

• μk = 0.2

• Acceleration due to gravity = 9.8 m / s²

• Initial velocity of the body = 0 (rest)

First let us write the net force acting on the block

⟹ F = mg sin θ - μk N

As per the diagram ,

⟹ N = mg cos θ

Hence ,

⟹ F = mg sin θ - μk mg cos θ

⟹ ma = mg sin θ - μk mg cos θ

⟹ a = g sin θ - μk g cos θ

Now let's substitute the given values in the above equation ,

⟹ a = 9.8 sin 30 - 0.2 x 9.8  cos  30

⟹ a = 9.8 x 1 /2 - 1.96 x  √3 /2

⟹ a =  4.9  - 0.98 x 1.73

⟹ a = 4.9 - 1.69

⟹ a = 3.21 m/s²

Now let's find the time taken by the body to reach the bottom of the plane

By applying second equation of motion ,

s = ut + at ² /2

here ,

• s = distance

• u = initial velocity

• t = time taken

• a = acceleration

⟹ s = at ² /2

⟹ t =  √ 2s / a

⟹ t =  √ 2 x 6.4 / 3.21

⟹ t =   √ 12.8 / 3.21

⟹ t =   √ 3.98

⟹ t = 1.9 s

The time taken by the body to reach the bottom is 1.9 s