Block A of mass 35 kg is resting on a frictionless floor. Another block B of mass 7 kg is resting on it as shown in the figure. The coefficient of friction between the blocks is 0.5, while kinetic friction is 0.4. If a force of 100 N is applied to block B, the acceleration of the block A will be : (g = 10m/s^2)
Answers
Given: Block A of mass 35 kg is resting on a friction less floor. Another block B of mass 7 kg is resting on it .
To find: The acceleration of the block A ?
Solution:
- Now we have given that coefficient of friction between the blocks is 0.5, while kinetic friction is 0.4.
f(s) ≤ μN
- Now Normal force will be:
N = mg = 7 x 10 = 70 N
f(s) ≤ μ(s) x 70
f(s) ≤ 0.5 x 70
f(s) ≤ 35 N
- So the force applied is greater than static friction so the body is in motion, then there is kinetic friction.
f(k) = μ(k) x 70
f(k) = 0.4 x 70
f(k) = 28 N
- Now on 35 kg block, only kinetic friction force will be applied.
f(1) = f(k) = 28 N
- So acceleration of mass 35 kg will be:
a = f(1) / 35 = 28 / 35
a = 0.8 m/s^2
Answer:
So the acceleration of the block A is 0.8 m/s^2.
Answer:
Given: Block A of mass 35 kg is resting on a friction less floor. Another block B of mass 7 kg is resting on it .
To find: The acceleration of the block A ?
Solution:
Now we have given that coefficient of friction between the blocks is 0.5, while kinetic friction is 0.4.
f(s) ≤ μN
Now Normal force will be:
N = mg = 7 x 10 = 70 N
f(s) ≤ μ(s) x 70
f(s) ≤ 0.5 x 70
f(s) ≤ 35 N
So the force applied is greater than static friction so the body is in motion, then there is kinetic friction.
f(k) = μ(k) x 70
f(k) = 0.4 x 70
f(k) = 28 N
Now on 35 kg block, only kinetic friction force will be applied.
f(1) = f(k) = 28 N
So acceleration of mass 35 kg will be:
a = f(1) / 35 = 28 / 35
a = 0.8 m/s^2
Answer:
So the acceleration of the block A is 0.8 m/s^2.