Question

a block of mass 10 kg is pulled with a constant speed on a rough horizontal plane with a force of 19.6 n. the co-officient of friction is....?

Answer 1

m=10kg,f=19.6N

Formula:f=Us(N)

Us=coefficient of friction,N=normal reaction

f=Us(m)(g)

19.6=Us(10)(9.8)

Us=19.6/(10)(9.8)

=2/10=0.2

Answer 2

Concept

Empirically it is found that for a body moving on a rough surface with constant speed under an applied force F experiences a friction force f whose magnitude is given by

                            $f = \mu_k N$,                   ...(1)

where $\mu_k$ is a constant, called the coefficient of kinetic friction,

$N$ is the normal force exerted by the surface on the body.

Given

• The mass of the body is 10kg
• The body is moving with constant speed.
• The applied force is 19.6N.

Find

The coefficient of kinetic friction.

Solution

Forces acting on the body

Four forces are acting on the body:

• The force of gravity in the vertical direction.
• The normal force, N, by the surface in the vertical direction.
• The applied force, F, in the horizontal direction.
• The force of friction, f, in the horizontal direction.

Friction force on the body

The body is moving at a constant speed on the plane surface so the net force, in the horizontal direction, acting on the body must be zero.

In the horizontal direction, as mentioned above, the magnitudes of the two forces: F, and f. So,

      F - f = 0

⇒   f = F.

The friction force on the body, thus, has a magnitude f = 19.6N.

The normal force acting on the body

The body is not moving in the vertical direction so the forces acting on the body in the vertical direction must balance. So,

      N - mg = 0

⇒   N = mg.

Calculate the coefficient of kinetic friction

Put the values of N and f in the equation (1)

      19.6 = $\mu_k$ mg

⇒      $\mu_k$ = (19.6N) / (10kg × 9.8m/s²)

⇒      $\mathbf{\mu_k}$ = 0.20.

The coefficient of kinetic friction is 0.20.

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