Question

Q.3 Consider a block of mass m= 10 kg kept on a smooth horizontal table as shown in the above figure.
If a horizontal force of 5 N acts on the block, the value of frictional force exerted by the table
on the block will be

Choose answer:

a)5 N
b)10 N
c)2N
d)Zero​

Answer 1

Answer:

a = 1/2

f = ma

f = 10×1/2

= 5 N

Answer 2

Answer:

The value of frictional force exerted by the table on the block will be 5N i.e.option(a).

Explanation:

The frictional force acting on the body is given as,

$f=\mu R$       (1)

Where,

f=frictional force acting on the body

μ=coefficient of friction

R=normal reaction acting on the body

And,

$\mu=\frac{a}{g}$        (2)

Where,

a=acceleration of the body

g=acceleration due to gravity=10m/s²

From the question we have,

Mass of the block(m)=10kg

The horizontal force acting on the block=5 N

The acceleration of the block is given as,

$a=\frac{5}{10} =\frac{1}{2} m/s^2$      (3)

By substituting the value of a and g in equation (2) we get;

$\mu=\frac{1}{2\times 10}=\frac{1}{20}$          (4)

And the normal reaction acting on the body is given as,

$R=mg$         (5)

By substituting the value of m and g in equation (5) we get;

$R=10\times 10=100N$      (6)

By using equations (4) and (6) in equation (1) we get;

$f=\frac{1}{20}\times 100=5N$

Hence, the value of frictional force exerted by the table on the block will be 5N i.e.option(a).