Answer:
Let f be force of friction on the blocks kept on inclined plane. T be tension in the string
For motion of block on the inclined plane in upward direction
T - m₁gsin40 - f = m₁a
f = μ m₁gcos40
For motion of hanging block on in downward direction
m₂g - T = m₂ a
Adding to cancel T
m₂g - - m₁gsin40 - μ m₁gcos40 = a ( m₁+m₂ )
a = g (m₂ - - m₁sin40 - μ m₁cos40) / ( m₁+m₂ )
Putting the values
a = 9.8 ( 4.75 - 2.12-1.045) / 7.6
2.04 m s⁻²
M₂ will go down and M₁ will go up with acceleration .
Explanation:
C
<span>Write at least 20 characters to explain it well.</span>
The equilibrant force of the two given forces is 14.14 N.
<h3 /><h3 /><h3>What is equilibrant force?</h3>
- This is a single force that balances other given forces.
The given parameters:
- First force, F₁ = 10 N
- Second force, F₂ = 10 N
- Angle between the forces, θ = 90⁰
The equilibrant force of the two given forces is calculated as follows;

Thus, the equilibrant force of the two given forces is 14.14 N.
Learn more about equilibrant force here: brainly.com/question/8045102
This statement would be FALSE.
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