Query Questions 13

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course Phy 241

October 7 around 1:30pmI just took out general phy and prin phy problems so it would be easier for you to grade, instead of scrolling through all the problems together.

I don't mind a little scrolling, but I certainly don't mind not having to do as much of it. What you've done is fine. It will also save you scrolling if you review these documents.

013. `query 13

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Question: `quniv phy 4.42 (11th edition 4.38) parachutist 55 kg with parachute, upward 620 N force. What are the weight and acceleration of parachutist?

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Your solution:

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Mass= 55.0 kg

Acceleration of gravity= 9.807 m/s^2

55.0kg*9.807m/s^2=539.38N (downward)

Net force:

-539.38N+620N=80.62N

Acceleration:

Acceleration=net force/mass

=80.62N/55.0kg=1.45m/s^2

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self-critique rating #$&*:

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Question: `qDescribe the free body diagram you drew.

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Your solution:

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I drew my free body diagram and showed the directions that the two forces were acting in. I declared up to be positive, therefore the combined weight of both was going to be downward and needed to be negative. The air resistance was upward, of +620N.

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confidence rating #$&*:

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Given Solution:

`aThe weight of the parachutist is 55 kg * 9.8 m/s^2 = 540 N, approx.. So the parachutist experiences a downward force of 540 N and an upward force of 620 N. Choosing upward as the positive direction the forces are -540 N and + 620 N, so the net force is

-540 + 620 N = 80 N.

Your free body diagram should clearly show these two forces, one acting upward and the other downward. The acceleration of the parachutist is a = Fnet / m = +80 N / (55 kg) = 1.4 m/s^2, approx..

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Self-critique (if necessary):

PERFECT!

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self-critique rating #$&*:

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Question: `quniv phy (4.34 10th edition) A fish hangs from a spring balance, which is in turn hung from the roof of an elevator. The balance reads 50 N when the elevator is accelerating at 2.45 m/s^2 in the upward direction.

What is the net force on the fish when the balance reads 50 N?

What is the true weight of the fish, under what circumstances will the balance read 30 N, and what will the balance read after the cable holding the fish breaks?

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Your solution:

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This wasn’t an assigned problem but I attempted it:

Net force on fish at 50N:

Fnet=ma

You have 2 forces acting on the fish: the 50N upward force from the cable and the downward force of “mg” due to gravity.

True weight of fish:

50N-mg=a

M=50N/(a+g)

=50N/(2.45m/s^2+9.807m/s^2)

=50N/12.257m/s^2

=4.08g

Balance will read 30N when:

Fnet=ma

30N-mg=a

30N-(4.08g)(9.807m/s^2)

= (-10N)

A= (-10N)/4.08g

=(-2.45m/s^2) so downward at 2.45m/s^2

When the cable holding the fish breaks, the balance will read:

0, simply because everything will be accelerating downward at the acceleration due to gravity, and therefore the balance won’t exert any force.

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confidence rating #$&*:

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Given Solution:

`a** Weight is force exerted by gravity.

Net force is Fnet = m * a. The forces acting on the fish are the 50 N upward force exerted by the cable and the downward force m g exerted by gravity.

So m a = 50 N - m g, which we solve for m to get

m = 50 N / (a + g) = 50 N / (2.45 m/s^2 + 9.8 m/s^2) = 50 N / 12.25 m/s^2 = 4 kg.

If the balance reads 30 N then

F_net = m a = 30 N - m g = 30 N - 4 kg * 9.8 m/s^2 = -9.2 N so

a = -9.2 N / (4 kg) = -2.3 m/s^2; i.e., the elevator is accelerating downward at 2.3 m/s^2.

If the cable breaks then the fish and everything else in the elevator will accelerate downward at 9.8 m/s^2. Net force will be -m g; net force is also Fbalance - m g. So

-m g = Fbalance - m g and we conclude that the balance exerts no force. So it reads 0. **

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Self-critique (if necessary):

I did everything right, however, I had a few more sig figs, which made the answer differ slightly. I also said “4.08g” when it should’ve been 4.08kg."

This looks great.

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