Query 132

course phy201

Query 13.2 013. `query 13

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Question: `qprin phy and gen phy problem 4.02 net force 265 N on bike and rider accelerates at 2.30 m/s^2, mass of bike and rider

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Your solution:

Fnet = 265N

a= 2.30m/s^2

fnet= m * a

265N= m * 2.30m/s^2

m= 115.22 kg

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Given Solution:

`aA force Fnet acting on mass m results in acceleration a, where a = Fnet / m. We are given Fnet and a, so we can solve the equation to find m.

Multiplying both sides by m we get

a * m = Fnet / m * m so

a * m = Fnet. Dividing both sides of this equation by a we have

m = Fnet / a = 265 N / (2.30 m/s^2) = 115 (kg m/s^2) / (m/s^2) = 115 kg.

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Self-critique (if necessary):

Ok, I got it

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Question: `qprin phy and gen phy problem 4.07 force to accelerate 7 g pellet to 125 m/s in .7 m barrel

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Your solution:

f= M * a

m = 7g * 1000g/1kg = 7000kg

Vf= 125m/s

'ds= .7m

'dt= 'ds/vave

'dt = .7m/ 62.5

'dt= .0112

a= 'dv/'dt

a= 125/.0112

a= 11160.71

F= 7000kg * 11160.71

f=78,124,970.0 newtons

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Given Solution:

`a** The initial velocity of the bullet is zero and the final velocity is 125 m/s. If we assume uniform acceleration (not necessarily the case but not a bad first approximation) the average velocity is (0 + 125 m/s) / 2 = 62.5 m/s and the time required for the trip down the barrel is .7 m / (62.5 m/s) = .011 sec, approx..

Acceleration is therefore rate of velocity change = `dv / `dt = (125 m/s - 0 m/s) / (.011 sec) = 11000 m/s^2, approx..

The force on the bullet is therefore F = m a = .007 kg * 11000 m/s^2 = 77 N approx. **

STUDENT COMMENT:

I did my answer a different way and came up with a number just off of this. I calculated 78 and this solution shows an answer of 77, but I am positive that I did my work right.

INSTRUCTOR RESPONSE:

The results of my numerical calculations are always to be regarded as 'fuzzy'. The calculations are done mentally and there is often no intent to be exact. This at the very least encourages students to do the arithmetic and think about significant figures for themselves.

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Self-critique (if necessary):

oooops I converted wrong......7g * 1kg/1000g= .007kg

f= .007kg * 11160.71 = 78.12 newtons

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Question: `qgen phy 4.08. A fish is being pulled upward. The breaking strength of the line holding the fish is 22 N. An acceleration of 2.5 m/s^2 breaks the line. What can we say about the mass of the fish?

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Your solution:

F= 22N

a= 2.5 m/s^2

f= m*a

m= f/a

m= 22N/2.5m/s^2

m= 8.8 kg

A big fish

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Given Solution:

`aThe fish is being pulled upward by the tension, downward by gravity. The net force on the fish is therefore equal to the tension in the line, minus the force exerted by gravity. In symbols, Fnet = T - m g, where m is the mass of the fish.

To accelerate a fish of mass m upward at 2.5 m/s^2 the net force must be Fnet = m a = m * 2.5 m/s^2. Combined with the preceding we have the condition

m * 2.5 m/s^2 = T - m g so that to provide this force we require

T = m * 2.5 m/s^2 + m g = m * 2.5 m/s^2 + m * 9.8 m/s^2 = m * 12.3 m/s^2.

We know that the line breaks, so the tension must exceed the 22 N breaking strength of the line. So T > 22 N. Thus

m * 12.3 m/s^2 > 22 N. Solving this inequality for m we get

m > 22 N / (12.3 m/s^2) = 22 kg m/s^2 / (12.3 m/s^2) = 1.8 kg.

The fish has a mass exceeding 1.8 kg.

STUDENT QUESTION

I had trouble understanding this question to begin with. I am a little confused on why the net force equals an acceleration of 12.3.

INSTRUCTOR RESPONSE

F_net = m a = m * 2.5 m/s^2, as expressed in the equation F_net = T - m g so that

m * 2.5 m/s^2 = T - m g.

It is the tension, not the net force, that ends up with an acceleration factor equal to 12.3 m/s^2:

T = F_net + m g = m * 2.5 m/s^2 + m * 9.8 m/s^2, which is where the 12.3 m/s^2 comes from.

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Self-critique (if necessary):

ok, so

f_net = T-mg

F= 22N

a= 2.5 m/s^2

F_net = T-m (9.8m/s^2 + 2.5m/s^2)

22N= T- m 12.3m/s^2

T+ 22N/12.3m/S^2 = m

m= 1.7886kg

??????????So the T does not really factor out of the equation it is just known that it is greater thatn or less than the Fnet??????????

First, check the slightly revised solution that follows. My answer to your question follows the revised solution. Note the use of capital M for the mass of the fish (lowercase m was used in the original given solution).

The fish is being pulled upward

by the tension, downward by gravity. 

The net force on the fish is therefore equal to the tension in the line,

minus the force exerted by gravity. 

In symbols, Fnet = T - M g, where

M is the mass of the fish.

style='mso-spacerun:yes'>  (We use capital M for the

mass of the fish to distinguish the symbol for mass from the symbol m for

meter).

 

To accelerate a fish of mass M upward at 2.5 m/s^2 the net

force must be Fnet = M a = M * 2.5 m/s^2. 

Combined with the preceding we have the condition

 

M * 2.5 m/s^2 = T -

M g so that to

provide this force we require

 

T = M * 2.5 m/s^2 +

M g = M * 2.5 m/s^2 +

M * 9.8 m/s^2 = M

* 12.3 m/s^2.

 

We know that the line breaks, so the tension must exceed the

22 N breaking strength of the line. 

So T > 22 N.

style='mso-spacerun:yes'>  Thus

 

M * 12.3 m/s^2 > 22 N.

style='mso-spacerun:yes'>  Solving this inequality for m we get

 

M > 22 N / (12.3 m/s^2) = 22 kg

m/s^2 / (12.3 m/s^2) = 1.8 kg.

 

The fish has a mass exceeding 1.8 kg.

STUDENT QUESTION

 I had trouble understanding this

question to begin with. I am a little confused on why the net force equals an

acceleration of 12.3.


INSTRUCTOR RESPONSE

F_net = M a = M * 2.5 m/s^2, as expressed

in the equation F_net = T - m g so that

It is the tension, not the net force, that

ends up with a factor of 12.3 m/s^2:

Nothing actually accelerates at 12.3

m/s^2, just as nothing in this system accelerates at 9.8 m/s^2. 

Thus the tension force is M * 2.5 m/s^2 +

M * 9.8 m/s^2 = M * 12.3 m/s^2.

Now in relation to your question:

Fnet is M * 2.5 m/s^2.

We know that T = M * 12.3 m/s^2, and that since the string breaks T is at least 22 N.

So M * 12.3 m/s^2 is at least 22 N, and M must be at least 22 N / 12.3 m/s^2 = 1.8 kg.

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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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Self-critique (if necessary):

m= 55kg

fnet= 620N

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Question: `qDescribe the free body diagram you drew.

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Your solution:

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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):

ok, so

55kg * 9.8m/s^2 = 540n

-540N Down + 620N= 80N

80N= 55kg * a

a = 80N/55kg

a= 1.45m/s^2

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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:

The balance is going in the opposite direction of the elevator because gravity is pulling on it as it hangs

so I think

Balance

F= 50N * 9.8m/s^2

F=-490N down

Fnet=50N-490N= 440N

440N= m * 2.45m/s^2

m= 179.59 kg

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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):

ok, so we use Fnet= t- mg

Fnet = 50N

T+50N= m * (9.8m/s^2+ 2.45m/s^2)

m= 50N/12.25m/s^2

m= 4.08kg

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Question: `qSTUDENT Self-critique rating #$&*

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Question: I had trouble with the problems involving tension in lines. For example the Fish prob.

Prob#9 A person yanks a fish out of water at 4.5 m/s^2 acceleration. His line is rated at 22 Newtons Max, His line breaks, What is the mass of the fish.

Here's what I did.

Sum of F = Fup + F down

-22 N = 4.5 m/s^2 * m(fish) - 9.8 m/s^2 * m(fish)

-22N = -5.3 m/s^2 m(fish)

m(fish) = 4.2 kg

I know its wrong, I just don't know what to do.I had the same problem with the elevator tension on problem 17.

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Your solution:

F_net= T- mg

T+ F_net = m (4.5m/s^2+9.8m/s^2)

22N= m 14.3m/s^2

m= 1.54kg

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Given Solution:

`a** Think in terms of net force.

The net force on the fish must be Fnet = m a = m * 4.5 m/s^2.

Net force is tension + weight = T - m g, assuming the upward direction is positive. So

T - m g = m a and

T = m a + m g. Factoring out m we have

T = m ( a + g ) so that

m = T / (a + g) = 22 N / (4.5 m/s^2 + 9.8 m/s^2) = 22 N / (14.3 m/s^2) = 1.8 kg, approx..

The same principles apply with the elevator. **

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Self-critique (if necessary):

ok, I understand, but i am still a little confused with the T where does it go???

T is at least 22 N, in order to break the string. So we replaced T with 22 N, in order to get the minimum mass of the fish.

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&#This looks very good. Let me know if you have any questions. &#