Question

2. In the given figure, F = 10 N and ma = mb = 10 kg, then calculate the force of
friction acting on the block B by ground in N.
A
u = 0.5 10 kg
10 N
H = 0.5
B 10 kg​

Answer 1

Answer:

We know that maximum friction is given by F=μ×N.

According to the given data, we have,

N=5×g=50

maximum friction force=μN=.5×50=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=Mf=20/10

a=2 m/s2

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

F−20=ma

F=30 newton

Explanation:

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Answer 2

F=30 newton 

Explanation:

F= μ x N is known is the formula to calculate the maximum friction.

We have, based on the information provided,

N=5×g=50

N=.550=25 Newton maximum friction force

The given friction force is 20 Newton, which is less than the maximum.

As a result, it's a static situation, which implies they'll both move at the same time.

The 10 kg Block's acceleration is determined by,

a=Mf=20/10

a=2 m/s2 

Now, if we apply Newton's second law to a 5 kilogramme block, we get:

F−20=ma

F=30 newton