Question

Coefficient of friction between 5 kg and 10 kg block is 0.5. If friction force between them is 20 N. What is the value of force being applied on 5 kg. The floor is frictionless

• A. 60 N
• B. 30 N
• C. 90 N
• D. 15 N

Answer 1

The correct option is B 30 N

We know that maximum friction is given by $F=\mu \times N$.
According to the given data, we have,
$N=5\times g=50$

maximum friction force=$\mu N=.5\times 50=25$ Newton

Given friction force is $20$ Newton which is less than maximum friction force.

Therefore it's static case which means both of them will move together.

Acceleration of 10 kg Block is given by,

$a=\frac{f}{M}=20/10$
$a=2$ m/s${}^2$

Now applying newton's second law on 5 kg block will give,

$F-20=ma$
$F=30$ newton