- #1

jtaz612

- 4

- 0

F(t) = (10.0 + 2.00 t2) Newtons

(a) If the mass starts from rest and the force acts for 5.00 sec, what is the resulting speed of the mass?

(b)Over the 5.00 sec period, how much work did the external force do on the mass?

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- Thread starter jtaz612
- Start date

- #1

jtaz612

- 4

- 0

F(t) = (10.0 + 2.00 t2) Newtons

(a) If the mass starts from rest and the force acts for 5.00 sec, what is the resulting speed of the mass?

(b)Over the 5.00 sec period, how much work did the external force do on the mass?

- #2

Nacer

- 10

- 0

Your attempts to solve it???

Regards,

Nacer.

Regards,

Nacer.

- #3

jtaz612

- 4

- 0

Oh yes my attempt:

You can determine the acceleration from using acc = Force/mass

or: a = ( 10.0 + 2.00 t2 ) / 10.00 = 1 + 0.2 t2

dv = a dt and add these, by intergating, to get the total final speed.

So: final speed = integral of ( a dt ) = integral of [ (1 + 0.2 t2 ) dt ] =

= ( t + (0.2/3) t3 ) eval from 0 to 5 = ( 5 + (0.2/3)*125) = 13.33 m/s

b) the work done is the change in KE, so in this case work is equal to final KE:

(1/2) m v2 = 1/2 * 10 * 13.332 = 888 Joules

Looks right?

- #4

denverdoc

- 963

- 0

- #5

jtaz612

- 4

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- #6

denverdoc

- 963

- 0

NP. good luck on your exam--if the work on it is as clear as your post, you'll do well.

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