Calculate Distance for Work Done by Force

It should be obvious what constitutes a complete solution. In any event, the OP should be able to work out what the answer is from what has been said so far.In summary, to find the distance a rock was lifted, you can use the equation W = FD, where W is the work done (in joules), F is the force applied (in newtons), and D is the distance the rock was lifted (in meters). In this problem, the force of 100 Newtons and the work of 150 joules were given, and by rearranging the equation to isolate D, we can find that the rock was lifted a distance of 1.5 meters. It is important to pay attention to the units of measurement in
  • #1
Jajo
5
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w = Force times Distance
w = FD

4. A force of 100 Newtons was necessry to lift a rock. A total of 150 joules of work was done. How far was the rock lifted??

Please i really need to understand how to do this problem, if anyone reply's please try to give step by steps
 
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  • #2
Jajo said:
w = Force times Distance
w = FD

4. A force of 100 Newtons was necessry to lift a rock. A total of 150 joules of work was done. How far was the rock lifted??

Please i really need to understand how to do this problem, if anyone reply's please try to give step by steps

Well, I'm not going to do it for you! You have an equation with three variables. You know two of them, and want to find the third. How do you think you should do this?
 
  • #3
i tried forever but i can't find a way to do this...so i have the force (100N) and i have the Work (150Joules) so i how can i get the distance
 
  • #4
Jajo said:
i tried forever but i can't find a way to do this...so i have the force (100N) and i have the Work (150Joules) so i how can i get the distance

You have the equation W=fd. Try rearranging this to make d the subject (i.e. to get d on its own). Can you do this?

[Hint: What does the equation look like if you divide both sides by f?]
 
  • #5
so i would divide d to both sides so it would be w/d = f right?
 
  • #6
Jajo said:
so i would divide d to both sides so it would be w/d = f right?
No; you want to find d.

Did you read this?
cristo said:
[Hint: What does the equation look like if you divide both sides by f?]
 
  • #7
so f = 1.5 meters or Newtons or joules
 
  • #8
Jajo said:
so f = 1.5 meters or Newtons or joules

No. The number is correct, but which one of those units do you mean, they are not equivalent.
 
  • #9
Ok i thought it was d...so the equation should look like this... W/F = D...so 150/100 = f...f = 1.5 meters or Newtons or Joules?
 
  • #10
Jajo said:
Ok i thought it was d...so the equation should look like this... W/F = D...so 150/100 = f...f = 1.5 meters or Newtons or Joules?

I made a mistake in my last post, you're correct in thinking it should be D, not f that is 1.5, but still it cannot be anyone of those units they are not equivalent. So which is it Newtons, meters or joules?
 
  • #11
"A force of 100 Newtons was necessry to lift a rock. A total of 150 joules of work was done. How far was the rock lifted??"

F=100N, right?
W=150J, right?

If W=FD, and you have F and W, but want to know D, set rearrange the equation to isolate the D.. W/F=D.. 100N/150J=1.5 meters.

Why 1.5 METERS? Well, Newtons is a metric unit, and Joules is a metric unit.. And the metric unit for distance is meter. Or.. you could expand the Newtons and joules..

Newton=kg*m/(s^2)
Joule=kg*(m^2)/(s^2)

Back to the problem.. [100*kg*m/(s^2)] / [150 kg*(m^2)/(s^2)]
That should result in 1.5, and all the units except meters cancel out, therefore, 1.5 meters.
 
  • #12
pugfug90 said:
"A force of 100 Newtons was necessry to lift a rock. A total of 150 joules of work was done. How far was the rock lifted??"

F=100N, right?
W=150J, right?

If W=FD, and you have F and W, but want to know D, set rearrange the equation to isolate the D.. W/F=D.. 100N/150J=1.5 meters.

Why 1.5 METERS? Well, Newtons is a metric unit, and Joules is a metric unit.. And the metric unit for distance is meter. Or.. you could expand the Newtons and joules..

Newton=kg*m/(s^2)
Joule=kg*(m^2)/(s^2)

Back to the problem.. [100*kg*m/(s^2)] / [150 kg*(m^2)/(s^2)]
That should result in 1.5, and all the units except meters cancel out, therefore, 1.5 meters.

Please don't provide complete solutions to homework problems.
 
  • #13
He and everyone else has pretty much solved it already, I'm just explaining "concepts" and "theory" so that he'll understand why the answer has to be in meters.
 
  • #14
pugfug90 said:
He and everyone else has pretty much solved it already, I'm just explaining "concepts" and "theory" so that he'll understand why the answer has to be in meters.

The OP had the correct numerical answer, however had not answered d_leet's question re the units. Besides, check out the PF guidelines: https://www.physicsforums.com/showthread.php?t=5374
guidelines said:
Under no circumstances should complete solutions be provided to a questioner, whether or not an attempt has been made.
 
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