Two masses mA = 1.2 kg and mB = 5.6 kg are on inclines and are connected together by a string as shown in (Figure 1) . The coefficient of kinetic friction between each mass and its incline is μk = 0.30.
If mA moves up, and mB moves down, determine their acceleration. Ignore masses of the (frictionless) pulley and the cord.
Express your answer to two significant figures and include the appropriate units. Enter positive value if the direction of the acceleration is in the direction of the velocity and negative value if the direction of the acceleration is in the direction opposite to the velocity.
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Physics homework, HELP!?
2016-05-30 5:25 am
回答 (2)
2016-05-30 6:13 am
✔ 最佳答案
Take g = 9.81 m/s²T : tension of the string
a : acceleration
When mA is moving up and mB is moving down :
Total force due to gravity towards the direction of movement, F(gravity)
= (mB g sin21°) - (mA g sin51°)
= 5.6 × 9.81 × sin21° - 1.2 × 9.81 × sin51° N
= 11 N
Total kinetic friction, F(friction)
= -(μk mB g cos21°) - (μk mA g cos51°)
= -0.30 × 5.6 × 9.81 × cos21° - 0.3 × 1.2 × 9.81 × cos51° N
= -18 N
Total force = F(gravity) + F(friction) = 11 - 18 N = -7 N < 0 N
Acceleration, a = 0.0 m/s²
2016-05-30 1:32 pm
The other poster is 98% correct: The net force is -7.1 N.
If the blocks were static, the acceleration would be zero.
But we're told "mA moves up, and mB moves down," so the masses decelerate.
Total system mass M = mA + mB = 6.8 kg
a = Fnet / M = -7.1N / 6.8kg = -1.0 m/s² ◄
where the sign means "against the direction of motion."
You may not need it in your answer.
Hope this helps!
If the blocks were static, the acceleration would be zero.
But we're told "mA moves up, and mB moves down," so the masses decelerate.
Total system mass M = mA + mB = 6.8 kg
a = Fnet / M = -7.1N / 6.8kg = -1.0 m/s² ◄
where the sign means "against the direction of motion."
You may not need it in your answer.
Hope this helps!
收錄日期: 2021-04-18 14:55:37
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