Math
posted by Samuel .
A track of mass 1000 kg travelling on a straight level road accelerates from a speed of 10 m/s to a speed of 30 m/s in 12 seconds. Given that the engine of the track exerts a constant pull of magnitude P N and that there is a constant resistance to motion of 200 N, find the value of P.

v changed from 10 to 30 in 12 sec.
So, the acceleration
a = 20/12 = 5/3 m/s^2
F = ma = 1000*5/3 = 1667 N
That is the net force, and since there is an opposing force of 200N,
P = 1667+200 = 1867 N
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