Question

A body of weight 200 N is being pulled by a force
of 200 N, which is being applied at an angle of 60°
to, the horizontal. The body is moving at this
instant with a velocity of 2 m/s. Find the velocity
of the body at the end of 15 s using impulse
momentum equation. The coefficient of friction
between the body and the surface is 0.25.​

Answer 1

Explanation:

According to question,

As shown in below figure

So,The force on the block long the horizontal is

F

x

=100cos30−f⋯(i)

f=μ×N⋯(ii)

and also,

200=N+100sin30

N=150N

Now,

substituting the above value in (ii)

f=150×μ

substituting the value of

′

f

′

in (i)

F

x

=100cos30−150×μ

Hence, the block moves with the constant velocity it means that there is no net force acting on it is horizontal direction,

F

x

=0

100cos30−150×μ=0

86.6025=150×μ

μ=0.57735