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Margaret [11]
3 years ago
13

A woman living in a third-story apartment is moving out. Rather than carrying everything down the stairs, she decides to pack he

r belongings into crates, attach a frictionless pulley to her balcony railing, and lower the crates by rope.
Required:
How hard must she pull on the horizontal end of the rope to lower a 49 kg crate at steady speed?
Physics
1 answer:
Flura [38]3 years ago
8 0

Answer:

T = 480.2N

Explanation:

In order to find the required force, you take into account that the sum of forces must be equal to zero if the object has a constant speed.

The forces on the boxes are:

T-Mg=0      (1)

T: tension of the rope

M: mass of the boxes 0= 49kg

g: gravitational acceleration = 9.8m/s^2

The pulley is frictionless, then, you can assume that the tension of the rope T, is equal to the force that the woman makes.

By using the equation (1) you obtain:

T=Mg=(49kg)(9.8m/s^2)=480.2N

The woman needs to pull the rope at 480.2N

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Answer:

|\vec r|=339.82\ m

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<u>Displacement </u>

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The monkey first makes a displacement given by (0.198 km,0°). The angle is 0 because it goes to the East, the zero-reference for angles. Thus the first displacement is

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|\vec r|=\sqrt{337.52^2+(-39.48)^2}=339.82\ m

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