Question

a body of mass m moves with velocity v on a surface whose coefficient of friction is mu=0.4 . If the body covers a distance S=8m before stopping , then value of v is


​

Answer 1

Explanation:

Mass of the block=5kg

Coeffecient of friction=0.2

external applied force, F=40N

The angle at which the force is applied=30degree

So the horizontal component of force=Fcos30=40×

2

3

=20

3

N

While the uertical component of the force acting in upward direction=Fsin30=40×

2

1

=20N

The normal reaction from the surface (N)=mg−Fsin30=50−20=30N

So the ualue of limiting friction=μN=0.2×30=6N

Hence the net horizontal force on the block=Fcos30=μN=20

3

N−6N=28.64N

The horizontal acceleration of the block=

m

Fcos30−μN

=

5

28.64

=5.73m/s

2