As you are trying to move a heavy box of mass m, you realize that it is too heavy for you to lift by yourself.There is no one ar
ound to help, so you attach an ideal pulley tothe box and a massless rope to the ceiling, which you wrap aroundthe pulley. You pull up on the rope to lift the box. Use g or themagnitude of the acceleration due to gravity and neglect frictionforces.
Part A What is the magnitude F of the upward force you must apply to the ropeto start raising the box with constant velocity?
Part A What is the magnitude F of the upward force you must apply to the ropeto start raising the box with constant velocity?
1 answer:
Answer: magnitude of applied force is FA = mg + F
Where F is the resultant force downward that the rope moves with
Explanation:
Force downwards F is,
F = FA - T
T is the upwards tension force on the rope
FA is the actual applied force in pulling the rope down.
Therefore, T = FA - F .....equ. (1)
For the box to move up with force ma ( it's mass times its acceleration upwards) upwards tension on the roap must exceed its own weight mg ( it's mass times acceleration due to gravity 9.8m/s^2)
Therefore, ma = T - mg
T = ma + mg ..... equ. (2)
Equating equ. 1 and 2
T = FA - F = ma + mg
Therefore FA = ma + mg + F
But at constant velocity a = 0
Magnitude of applied force becomes
FA = mg + F
See image below
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13.0m/s
1.2m/s
Explanation:
Given parameters:
Initial speed of the body = 7.1m/s
time taken = 2.23s
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Unknown:
Final speed = ?
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Acceleration is the rate of change of velocity with time.
a =
a = acceleration
V = final speed
U = initial speed
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2.64 =
V - 7.1 = 5.9 expression 1
V = 5.9 + 7.1 = 13.0m/s
B
Using the same parameters, the speed after a uniform deceleration of -2.64m/s², the negative sign implies deceleration;
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V - 7.1 = -5.9
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Answer:
the magnitude of the work done by the two blocks is the same.
Explanation:
The work done by block a on block b is given by:
where Fa is the force exerted by block a on block b, and d is the distance they cover.
The work done by block b on block a is given by:
where Fb is the force exerted by block b on block a, and d is still the distance they cover.
For Newton's third law, the force exerted by block a on block b is equal to the force exerted by block b on block a, therefore
and so