How Much Work Is Done on a Mass by a Time-Varying Force?

[XwakeriderX]

Homework Statement

A time-varying horizontal force is given by,
F(t) = (1.00 t + 0.250 t^2) Newtons
The force is applied from 0 to 5.00 seconds on a 10.0 kg mass which is initially at rest but is allowed to slide freely with only this one horizontal force acting on it. How much work is done on the mass by force F during the 5 seconds?

Homework Equations

W=Integral (from Xi to Xf) F_xdx

The Attempt at a Solution

W=(t^2)/2 + (t^3)/12 from 0 to 5
W=22.917

That is what i got from integrating.. But the answer is 26.3

 

[tiny-tim]

Hi XwakeriderX!

(have an integral: ∫ and try using the X² and X₂ icons just above the Reply box )

A time-varying horizontal force is given by,
F(t) = (1.00 t + 0.250 t^2) Newtons
The force is applied from 0 to 5.00 seconds on a 10.0 kg mass which is initially at rest but is allowed to slide freely with only this one horizontal force acting on it. How much work is done on the mass by force F during the 5 seconds?
…
W=Integral (from Xi to Xf) Fxdx
…
W=(t^2)/2 + (t^3)/12 from 0 to 5

No, that's ∫ F dt, and (as you say) you need ∫ F dx …

try again ​

 

[XwakeriderX]

Ah okay i got it! i used impulse then apllied W=Pf^2/2M

 

[tiny-tim]

Ah okay i got it! i used impulse then apllied W=Pf^2/2M

remember of course that this isn't a general formula, it works here only because v_i = 0 …

generally ∆W = ((v_i + P)² - v_i²)/2m = (P² + 2v_iP)/2m

 

[rwisz]

Joules.

I would like to point out that your approach and calculation are correct, but there may be some rounding errors or differences in the units used that could result in a slightly different answer. It is also possible that the given answer of 26.3 Joules is an approximation or rounded value. In any case, your understanding and application of the relevant equations and concepts is commendable. Keep up the good work.