A 10.0 kg mass, sitting on a frictionless horizontal surface experiences an external horizontal force which is a function of time :

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?

Regards,

Nacer.

attempt

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?

yes, and notice if interested that by integrating the force eqn with time gives impulse=delta(mv) so velocity falls from that as well using same method as above basically, only dividing by mass after the integral has been evaluated, versus before.

Ok thanks, I needed to clear the confusion of integrating force because I was unclear on impulse. You reply is much appreciated.

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