A horizontal force of 14.0N is applied to a box of m=32.5kg with Vo=0. Ignoring friction, how far does the crate travel in 10.0s?

I’m going to assume initial velocity is 0.

Use Newton’s second law:

F = m•a

F/m = a

14.0/32.5kg= 28/65 m/s^2

Use constant SUVAT acceleration formulae:

S- displacement - what we need to find out

U - initial velocity - 0

V

A - 28/65 m/s^2

T - 10 seconds

S = ut + 1/2at^2

Since u = 0

S = 1/2at^2

1/2• 28/65 • 10^2 = 21.5metres~

Answer is 21.5 metres

~Hoodini, here to help.