Blocks X and Y are attached to each other by a light rope Y х
and can slide along a horizontal, rough surface. Block X has
F20p
a mass of 10 kg and block Y a mass of 5 kg. An applied force of 36 N (night) acts on block X
Suppose the magnitude of the force of friction on blocks X and Y are 8N and 4N respectively. The
magnitude of tension in the string between the blocks is
(A) 36 N
(B) 4N
(C) 12 N
(D) 18 N
Answers
Answered by
4
The magnitude of tension is T = 12 N
Explanation:
We are given that:
- Mass of block X = 10 Kg
- Mass of block Y = 5 Kg
- Applied force = 36 N
- Magnitude of force on X block = 8N
- Magnitude of force on Y block = 4 N
Solution:
Net force=36-8-4=24N
Total mass=10+5 =15kg
Acceleration of the system=24/15=1.6m/s2
For tension in the string
Forces considered on mass 10kg
Force applied-frictional force on X-tension=mass of X × acceleration
36-8-T=10×1.6
T = 12 N
Answered by
0
Answer:
The magnitude of tension is T = 12 N
Explanation:
We are given that:
Mass of block X = 10 Kg
Mass of block Y = 5 Kg
Applied force = 36 N
Magnitude of force on X block = 8N
Magnitude of force on Y block = 4 N
Solution:
Net force=36-8-4=24N
Total mass=10+5 =15kg
Acceleration of the system=24/15=1.6m/s2
For tension in the string
Forces considered on mass 10kg
Force applied-frictional force on X-tension=mass of X × acceleration
36-8-T=10×1.6
T = 12 N
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