Help: Potential Energy probs

  • #1
Hey for anyone up this late, im having some problems with three questions. all these are to be solved using K_1+U_1=K_2+U_2

1) A mail bag with a mass of 115kg is suspended by a vertical rope of length 7.00m .A)What horizontal force is necessary to hold the bag in a position displaced sideways a distance 3.00m from its initial position?
Take free fall acceleration to be = 9.80m/s^2 . B) How much work is done by the worker in moving the bag to this position?
Take free fall acceleration to be = 9.80m/s^2 .

Part A was easy and i figured it to be 535N but part B is killing me. I tried using W=F*s but it came out wrong and your hinted to use g=9.8m/s^2....any ideas or tips/hints?

2)A spring of negligible mass has force constant = 1800N/m. A) How far must the spring be compressed for an amount 3.20J of potential energy to be stored in it? B) You place the spring vertically with one end on the floor. You then drop a book of mass 1.50kg onto it from a height of 0.900m above the top of the spring. Find the maximum distance the spring will be compressed. Take the free fall acceleration to be = 9.80m/s^2.

Again part A was easy and i found it to be 5.96×10^−2 m but part B is being troublesome. I tried using W=(1/2)K(x_2)^2-(1/2)K(x_1)^2 and made W=mgh which came out to be 13.23J and set this equal to
(1/2)K(x_2)^2-(1/2)K(x_1)^2 and solved for the delta x but got it wrong. Any ides on where my mistakes are?

3)A wooden rod of negligible mass and length 75.0cm is pivoted about a horizontal axis through its center. A white rat with mass 0.510kg clings to one end of the stick, and a mouse with mass 0.150kg clings to the other end. The system is released from rest with the rod horizontal. If the animals can manage to hold on, what are their speeds as the rod swings through a vertical position? Take free fall acceleration to be = 9.80m/s^2

Ok, this one has had me stumped for a few hours now and iv'e tried a couple of different methods. I tried K_1+U_1=K_2+U_2 making K_1 and U_2 equal to zero. eventually i came to the expression of V_2=sqrt(2gr) and used 75cm/2 as r but got the wrong answer. Then i tried substracting the two masses to have an equivalent system where there is only one mass released form rest but agian that took me no where. Any ideas here?
 

Answers and Replies

  • #2
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For Q(1) part B, can you draw a diagram to see the actual path over which the mail bag is displaced by the force? Also, do you know how to compute work done if the force (magnitude or direction) is changing along the displacement path?
 
  • #3
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For question 2, part (B): Your idea seems correct. The decrease in potential energy is equal to the maximum elastic potential energy. You could be making a computational mistake. The spring is initially in free state by the way.
 
  • #4
ooooo are you talking about work along a curved path where W=int(F*dl) ?
 
  • #5
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cukitas2001 said:
ooooo are you talking about work along a curved path where W=int(F*dl) ?

[itex]dW = \vec{F}*\vec{dl}[/itex]
 
  • #6
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Your answer for Q. 3 is correct if you have ONE mass at a distance of r from the pivot. But here you have two, so you have take into account the kinetic and potential energies of both.
 
  • #7
where do i start for the dW=F*dl, im looking at some examples in the book and its not making any sense. ima gonna try something on q3 now that u mention that
 
  • #8
on q3 the only answer i got was 2.71m/s, which is wrong and either i get v=sqrt(2gh) or end up with an equality with zero on one side and not being able to solve for v
 
  • #9
finished on q1. On q3 how can i combine the effects of boths mass' kinetic and potential energies? Could i just have something like
(K_1a+K_1b)+(U_1a+U_1b)=(K_2a+K_2b)+(U_2a+U_2b) where a and b are the masses respectively?
 
  • #10
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cukitas2001 said:
on q3 the only answer i got was 2.71m/s, which is wrong and either i get v=sqrt(2gh) or end up with an equality with zero on one side and not being able to solve for v

The rat goes down and the mouse goes up :smile:

Also, the speeds are same in this case ([itex]v=r\omega[/itex] and [itex]\omega[/itex] and [itex]r = (1/2)length[/itex] are both same for the rat and mouse).
 
  • #11
omega is angular velocity right? that's not till like 5 chapters :-/ ....dont i need time to find omega?
 
  • #12
Any ideas on q2 part b i know its something simple im missing
 
  • #13
Gokul43201
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For Q1 part B, you don't need to integrate anything.

HINT : All you need is the work-energy theorem (do you know this?) and Pythagoras.
 
  • #14
Gokul43201
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cukitas2001 said:
Any ideas on q2 part b i know its something simple im missing
All you need to do is use the equation you've written down in post #9 (the energy conservation equation), and apply it to the initial and final states of all parts of the system (system = book + spring).
 
Last edited:
  • #15
Gokul43201
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cukitas2001 said:
finished on q1. On q3 how can i combine the effects of boths mass' kinetic and potential energies? Could i just have something like
(K_1a+K_1b)+(U_1a+U_1b)=(K_2a+K_2b)+(U_2a+U_2b) where a and b are the masses respectively?
Yes, ,this is exactly what you want to use.
 
  • #16
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Gokul43201 said:
For Q1 part B, you don't need to integrate anything.

HINT : All you need is the work-energy theorem (do you know this?) and Pythagoras.

Yea that's a better idea.
 
  • #17
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cukitas2001 said:
omega is angular velocity right? that's not till like 5 chapters :-/ ....dont i need time to find omega?

All you need to know (just for the sake of understanding this problem right now) is that angular speed of both the rat and the mouse is the same because they are attached to the same rigid rod. This is true irrespective of the distance they are situated from the center. But if the distance is same, then the speeds are same (the velocities are opposite) because [itex]\omega = v/r[/itex] is constant. Here v is the speed and r is the distance from the axis of rotation. In your case, r = half the length of the rod. Is it clear now?
 

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