My answer: Solutions to Physics Problems: Pendulums, Pulleys, Friction, and Work

[:::JMANN:::]

1. A 1N pendulum bob is held at an angle theta from the vertical by a 2N horizontal force. Find the tension in the string supporting the pendulum bob.
My answer: 2.236N
Ty=mg=1N Tx=2N, T=sqr((2N)^2 + (1N)^2)=sqr(5*N^2)=2.236N
Without knowing the angle or length I don't know how else to solve this one.

4. Two blocks weighing 250N and 350N respectively are connect by a string that passes over a massless pulleye. Compute the tension in the string.
My answer: 291.4N (this 2T, T=145.7N)
I am assuming that the blocks would be moving due to the difference in weight. So, M=35.71kg and m=25.51kg
(M+m)a=Mg (35.71+25.51)kg*a=35.71kg(9.8m/s^2)
(61.22kg)a=349.96N a=349.96N/61.22kg a=5.72m/s^2
T=m*a=(35.71kg)(9.8-5.72)m/s^2 T=145.7N
In the book it stated that the magnitude of the rope tension on a pulleye system like this was 2T so it would be 291.4N, but that 2T is when masses are equal, so would it just be 145.7N?

6. At what angle should the roadway on a curve with a 50m radius be banked to allow cars to negotiate the curve at 12m/s even if the roadway is icy( and friction force is 0)?
My answer: @=17.09*
V=sqr(Rg(sin@+Us*cos@) if Us = 0 then, V=sqr(Rg(sin@))
12m/s=sqr(Rg(sin@)) 144(m/s)^2=Rg(sin@)
144(m/s)^2=(50m)(9.8m/s^2)(sin@)
144(m/s)^2=490(m/s)^2(sin@)
144/490=sin@ 0.294=sin@ @=sin^-1(0.294)=17.09*

7. A 5.0kg cart is moving horizontally at 6.0m/s. In order to change its speed to 10.0m/s, what must the net work done on the cart be?
My answer: 160J
change in K=W=Kf-Ki
1/2mVf^2 - 1/2mVo^2 1/2(5.0kg)(10.0m/s)^2 - 1/2(5.0kg)(6.0m/s)^2
2.5kg(100.0m^2/s^2) - 2.5kg(36.0m^2/s^2) 250J-90J= 160J

8. A person holds an 80N weight, 2 meters above the floor for 30 seconds, what is the power required in watts to do this?
My answer: 4800watts
Wg= -mgd Wg= -80N*2m Wg=-160J Wf=-Wg=160J
1watt=1J/s 30*160J/s=4800J/s=4800watts

10. The speed of a 4.0N hockey puck, sliding on the ice surface decreases at the rate of 0.61m/s^2. What is the coefficient of friction between the puck and the ice?
My answer: Uk=0.062
a=-0.61m/s^2 a=-Uk*g -0.61m/s^2= -Uk(9.8m/s^2)
-Uk= (-0.61m/s^2)/(9.8m/s^2)=-0.062 Uk=0.062

I think that I did these problems right but you never know, that's why I'd appreciate any feed back. #2,3, and 5 I am still working on and are posted in another thread, and #9 was ommited b/c it's a duplicate. Thanks much.

 

[Andrew Mason]

1. A 1N pendulum bob is held at an angle theta from the vertical by a 2N horizontal force. Find the tension in the string supporting the pendulum bob.
My answer: 2.236N
Ty=mg=1N Tx=2N, T=sqr((2N)^2 + (1N)^2)=sqr(5*N^2)=2.236N
Without knowing the angle or length I don't know how else to solve this one.

To find the angle, note that $T\sin\theta = 2$ and $T\cos\theta = 1$ so $\theta = arctan(2) = 63.4 \text{deg}$. T = 2/sin(63.4) = 2.236N

4. Two blocks weighing 250N and 350N respectively are connect by a string that passes over a massless pulleye. Compute the tension in the string.
My answer: 291.4N (this 2T, T=145.7N)
I am assuming that the blocks would be moving due to the difference in weight. So, M=35.71kg and m=25.51kg
(M+m)a=Mg (35.71+25.51)kg*a=35.71kg(9.8m/s^2)
(61.22kg)a=349.96N a=349.96N/61.22kg a=5.72m/s^2
T=m*a=(35.71kg)(9.8-5.72)m/s^2 T=145.7N
In the book it stated that the magnitude of the rope tension on a pulleye system like this was 2T so it would be 291.4N, but that 2T is when masses are equal, so would it just be 145.7N?

You should work out a solution algebraically and then plug in values. It is easier to follow, less prone to error and usually reduces to something simpler. For example, you don't need to determine masses here:

$$T - m_1g = m_1a$$

$$m_2g - T = m_2a$$

$$m_2g - m_1g = (m_1 + m_2)a$$

$$T = m_1g + m_1(m_2 - m_1)g/(m_1 + m_2) = m_1g(1 + (m_2 - m_1)/(m_1 + m_2)) = 250(1+(100/600)) = 292 N$$

6. At what angle should the roadway on a curve with a 50m radius be banked to allow cars to negotiate the curve at 12m/s even if the roadway is icy( and friction force is 0)?
My answer: @=17.09*
V=sqr(Rg(sin@+Us*cos@) if Us = 0 then, V=sqr(Rg(sin@))
12m/s=sqr(Rg(sin@)) 144(m/s)^2=Rg(sin@)
144(m/s)^2=(50m)(9.8m/s^2)(sin@)
144(m/s)^2=490(m/s)^2(sin@)
144/490=sin@ 0.294=sin@ @=sin^-1(0.294)=17.09*

I think you will find that $v^2 = rg\tan\theta$

7. A 5.0kg cart is moving horizontally at 6.0m/s. In order to change its speed to 10.0m/s, what must the net work done on the cart be?
My answer: 160J
change in K=W=Kf-Ki
1/2mVf^2 - 1/2mVo^2 1/2(5.0kg)(10.0m/s)^2 - 1/2(5.0kg)(6.0m/s)^2
2.5kg(100.0m^2/s^2) - 2.5kg(36.0m^2/s^2) 250J-90J= 160J

Again, an algebraic solution is much simpler:

$$W = \frac{1}{2}m(v_f^2-v_i^2) = .5*5*(64) = 160 J$$

8. A person holds an 80N weight, 2 meters above the floor for 30 seconds, what is the power required in watts to do this?
My answer: 4800watts
Wg= -mgd Wg= -80N*2m Wg=-160J Wf=-Wg=160J
1watt=1J/s 30*160J/s=4800J/s=4800watts

Does it take any more energy to hold it at 10 m? Remember, you are just holding it still.

10. The speed of a 4.0N hockey puck, sliding on the ice surface decreases at the rate of 0.61m/s^2. What is the coefficient of friction between the puck and the ice?
My answer: Uk=0.062
a=-0.61m/s^2 a=-Uk*g -0.61m/s^2= -Uk(9.8m/s^2)
-Uk= (-0.61m/s^2)/(9.8m/s^2)=-0.062 Uk=0.062

You are plugging in numbers before reducing to simplest form:

$$\mu_k = a/g = .61/9.8 = .062$$

AM

 

[:::JMANN:::]

Thanks man for looking that over. So I do have the amswers to all of the except the power question then?

 

[:::JMANN:::]

Quote:
8. A person holds an 80N weight, 2 meters above the floor for 30 seconds, what is the power required in watts to do this?
My answer: 4800watts
Wg= -mgd Wg= -80N*2m Wg=-160J Wf=-Wg=160J
1watt=1J/s 30*160J/s=4800J/s=4800watts

Does it take any more energy to hold it at 10 m? Remember, you are just holding it still.


If the Wg=-mgd and d is 2M, my answer would make sense, only if the object is being lifted up to 2M. If the height of 2M is the starting point, then d still has to be relavent to the equation right? Increasing the height to 10M would require more work, which would mean more power correct? Or is the height simply a referrence point and Wg= 0 given that there is no change in kenetic energy? I'm confused I think.

 

[Andrew Mason]

Quote:
8. A person holds an 80N weight, 2 meters above the floor for 30 seconds, what is the power required in watts to do this?
My answer: 4800watts
Wg= -mgd Wg= -80N*2m Wg=-160J Wf=-Wg=160J
1watt=1J/s 30*160J/s=4800J/s=4800watts

Does it take any more energy to hold it at 10 m? Remember, you are just holding it still.If the Wg=-mgd and d is 2M, my answer would make sense, only if the object is being lifted up to 2M. If the height of 2M is the starting point, then d still has to be relavent to the equation right? Increasing the height to 10M would require more work, which would mean more power correct? Or is the height simply a referrence point and Wg= 0 given that there is no change in kenetic energy? I'm confused I think.

Work is the product of force and a displacement: $W = \vec F \cdot \vec d$. If you hold something still, there is no displacement so the work is: ?

Look at it from the point of view of potential and kinetic energy. Does the potential or kinetic energy change? If you make a support and hang it there, do you have to continually supply energy to keep it hanging in the same spot?

AM

 

[Andrew Mason]

Thanks man for looking that over. So I do have the amswers to all of the except the power question then?

You need to correct 4. and 6. as well.

AM

 

[:::JMANN:::]

okay, for #8, there's no change in kenetic energy, so it would be 0 work, so there's no watts if there's no work.

#4 I got 291.667N

# I got 16.37*

 

[Andrew Mason]

okay, for #8, there's no change in kenetic energy, so it would be 0 work, so there's no watts if there's no work.

#4 I got 291.667N

# I got 16.37*

Apart from the significant figures, (the correct answer is 292) your answer is right. Are you now sure that it is T and not 2T?

Your answer for 6 is correct subject to the correct significant figures.

AM

 

[:::JMANN:::]

The book says on a pulley system that the tension is 2T. Follow the equation for acceleration on the blocks due to mass, the original T=145.7N then 2T=291.4N roughly 292N like your equation resulted. As far as significant figures, I don't think that the professor will be that picky, as long as we are reasonably close.

I am going to class to turn it in. I appreciate all the help that I recieved. I'm hoping to do well.

 

[Andrew Mason]

The book says on a pulley system that the tension is 2T. Follow the equation for acceleration on the blocks due to mass, the original T=145.7N then 2T=291.4N roughly 292N like your equation resulted. As far as significant figures, I don't think that the professor will be that picky, as long as we are reasonably close.

I am going to class to turn it in. I appreciate all the help that I recieved. I'm hoping to do well.

But the tension is T. T is 292 N. The force on the pulley is 2T, but the tension force on each mass is only T. Don't try to memorize formulae. Learn the principles and you will be able to figure these things out yourself.

And, BTW, you cannot have a result that is more accurate than the information given, which is to 3 significant figures. This is very important in doing any experimental work. You should learn now how and when to use sigfigs so you will not overstate your conclusions when you make your first Earth shattering discovery and publish the results.

AM

 

[:::JMANN:::]

But the tension is T. T is 292 N. The force on the pulley is 2T, but the tension force on each mass is only T. Don't try to memorize formulae. Learn the principles and you will be able to figure these things out yourself.

And, BTW, you cannot have a result that is more accurate than the information given, which is to 3 significant figures. This is very important in doing any experimental work. You should learn now how and when to use sigfigs so you will not overstate your conclusions when you make your first Earth shattering discovery and publish the results.

AM

Well I turned in the test, and so far I got all of them right, with the exception of #2, the rocket fuel question, which we didn't go over, so I don't know the answer, but I don't think that I got it right. The Tnet=292N like we got for the final answer, but for the way I did it, Tnet for the pulley system is 2T of my tension for the equation I used. Either way the answer was right, how it was derived won't effect that.

 

[Andrew Mason]

The Tnet=292N like we got for the final answer, but for the way I did it, Tnet for the pulley system is 2T of my tension for the equation I used. Either way the answer was right, how it was derived won't effect that.

Yes it will. The point here is not to get the right answer but to understand the physics. If you think 2T = 292N then the answer is wrong. The tension in the string is T. What is Tnet?

AM

 

[:::JMANN:::]

Yes it will. The point here is not to get the right answer but to understand the physics. If you think 2T = 292N then the answer is wrong. The tension in the string is T. What is Tnet?

AM

for this particular problem 2T is actually not 292N T=292N, however for the problem that I used, the equation was for an accerating block on a vertical plane, pulling a block on a horizontal plane, where T=146N. Then if the it were a hanging pulleye system as describe, T= 2T in respect to the horizontal equation. I didn't know how to solve the equation with a new formula, so I used the one I knew, found T=146, found 2T=292N, so Tnet in the vertical pulleye system is equal to 2T of the horizontal, so yes 2T=292N, just not for the problem directly from the test. We both got the correct answer of 292N, i just went about it another way. Thanks again for the help, I greatly appreciate it.

 

[Andrew Mason]

for this particular problem 2T is actually not 292N T=292N, however for the problem that I used, the equation was for an accerating block on a vertical plane, pulling a block on a horizontal plane, where T=146N. Then if the it were a hanging pulleye system as describe, T= 2T in respect to the horizontal equation. I didn't know how to solve the equation with a new formula, so I used the one I knew, found T=146, found 2T=292N, so Tnet in the vertical pulleye system is equal to 2T of the horizontal, so yes 2T=292N, just not for the problem directly from the test. We both got the correct answer of 292N, i just went about it another way. Thanks again for the help, I greatly appreciate it.

You should know how to solve the equation using the right analysis, not copying something from another problem that you don't understand. I am not trying to make it difficult for you. I am trying to make it easier. There is no force that is equal to 146 N in this problem. The tension in the string is 292 N. 2T is not 292 N. 2T is 584 N.

If you want to determine the force on the PULLEY, you must use 2T, because there the rope is doubled over the pulley (one T on each side = 2T). But if you want to determine the force on ONE MASS, you must use T, because there is only a single rope applied to the mass.

AM

 

[:::JMANN:::]

I realize that there is no tension = 146N in this problem. I took the values from this problem and plugged them into another problem. T from this problem would = 2T from the other problem, that's how I got 292N in the first place. I understand that I didn't really use the correct formula, but at the time that's what I found and used. The answer is 292N. Thats what you got that's what I got. The professor went over the problem the way you did it, though I am still having trouble grasping the concept behind the way you guys did it. I guess I'm having a hard time graspin all the physics concepts.