Physics Help Forum Deriving equations (3 part question!)

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 Oct 7th 2017, 10:28 PM #1 Junior Member   Join Date: Oct 2017 Posts: 11 Deriving equations (3 part question!) I need help with all the following and will post under each what I believe the answer is. Thank you in advance! A body of mass m accelerates uniformly from rest to a speed of V in total time T. (a) Mathematically derive an equation for the work done on the body as a function of time T. W= ∫FD and F=ma -> F=m(dv/dt) so ∫F(t)/m dt ? (b)Now derive an equation for instantaneous power delivered to the object Power = work/time not really sure where to go for this one. (c) Calculate the power at the end of 5sec delivered to a body of 1500N which accelerates 2m/s in 10s. ½mv²= KE = work P= work/ 20 again, not really sure. Any help appreciated
Oct 8th 2017, 06:49 AM   #2
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 Originally Posted by alex34 I need help with all the following and will post under each what I believe the answer is. Thank you in advance! A body of mass m accelerates uniformly from rest to a speed of V in total time T. (a) Mathematically derive an equation for the work done on the body as a function of time T. W= ∫FD and F=ma -> F=m(dv/dt) so ∫F(t)/m dt ?
No, with F= m(dv/dt), F(m)= (dv/dt) so ∫F(t)/m dt= ∫(dv/dt)dt= ∫dv= v. that gives v, not W.

Your "W= ∫FD" should be "W= ∫F dx". Then v= dx/dt so that dx= vdt. That can be written as "W= ∫ F v dt". Now, with F= ma= m(dv/dt) "W= ∫F v dt= ∫m(dv/dt) v dt= ∫mv dv. For constant m, that is (1/2)mv^2.

 (b)Now derive an equation for instantaneous power delivered to the object Power = work/time not really sure where to go for this one.
"Power= work/time" only for constant work and time. More generally, power= the derivative of work with respect to time.

 (c) Calculate the power at the end of 5sec delivered to a body of 1500N which accelerates 2m/s in 10s. ½mv²= KE = work P= work/ 20 again, not really sure. Any help appreciated
How fast was the energy being delivered to the body at t= 5 sec?

Oct 8th 2017, 09:41 AM   #3
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For part (a) wouldn't the work equation just stop at W = ∫Fvdt ?
 Now, with F= ma= m(dv/dt) "W= ∫F v dt= ∫m(dv/dt) v dt= ∫mv dv. For constant m, that is (1/2)mv^2.
If i need to derive it as a function of time t the integral would have to be with respect to t right? Or are you just simply showing me the process more.

Part (b) makes sense I should have known power = dw/dt.

part (c)
 How fast was the energy being delivered to the body at t= 5 sec?
It would be delivered at 10m/s. Not sure where to keep going from here. Thanks for all the assistance today.

 Oct 8th 2017, 02:46 PM #4 Junior Member   Join Date: Oct 2017 Posts: 11 Part (a) So shouldn't I leave it at W= ∫ F v dt because I need an equation a function done on time T so I would need to integrate with respect to t? Part(b) Power = dw/dt. I should have known this thank you. Part(c) The speed would be delivered at 10m/s. So to solve this would I take the derivate of work with respect to time using the equation for part a ? Thanks!
 Oct 8th 2017, 03:03 PM #5 Senior Member   Join Date: Aug 2010 Posts: 369 I don't know what you mean by speed being "delivered".
 Oct 8th 2017, 03:22 PM #6 Junior Member   Join Date: Oct 2017 Posts: 11 Sorry, I meant at t = 5 the energy would be delivered at a rate of v = 10m/s. ?
Oct 9th 2017, 07:17 AM   #7
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 Originally Posted by alex34 Part (a) So shouldn't I leave it at W= ∫ F v dt because I need an equation a function done on time T so I would need to integrate with respect to t?
This would work only if v is constant, but it's not. For constant acceleration the position of the body as a function of time is:

$\displaystyle x = \frac 1 2 a t^2$

Therefore:

$\displaystyle dx = a t dt$

So you can manipulate the integral of Fdx using:

$\displaystyle W = \int F dx = \int F a t dt = \frac 1 2 F a t^2$

Can you take it from here?

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