Question

A sled of 6 kg mass is moving along a smooth, horizontal ice surface with a velocity of Vo. A force of 36 N is applied to the sled in it direction of motion, increasing its velocity to 2Vo when it moves 10 m. find the sleds original velocity, Vo and the length of time the force acted.

Answer 1

36 = 6 a
a = 6 = change in velocity / time = Vo/t
d = 10 meters = average speed * t = 1.5 Vo t
so Vo = 10 / 1.5 t
so
6 = ( 10 /1.5 t) / t
6 t = 10/1.5t
9 t^2 = 10
t = 1.05 seconds
Vo = 10/1.5 t = 6.35m/s
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check
Vo = 6.35
2 Vo = 12.7
1.5 Vo = 9.52
t = 10 m / 9.52 m/s = 1.05 s
a = dv/dt = 6.35 / 1.05 = 6.05 m/s^2
F = m a = 6 *6.05 = 36.3 Newtons, close enough

Answer 2

How’d you find time ?