Acrate with mass 32.5 kg initially at rest on a warehouse floor is acted on by a net horizontal force of 140 n.
a)what acceleration is produced?
b)how far does the crate travel in 10.0 s?
c)what is its speed at the end of
10.0 s?

Newton's second law of motion may be mathematically expressed as Fnet = m x a.

a. a = F / m. Substituting the known values,
                              a = 140 N / 32.5 kg = 4.3 m/s²
b. d is calculated through the equation, d = Vit + 0.5at². Given that Vi is zero, the distance is,
                                    d = 0.5(4.3 m/s²)(10 s)² = 215 m
c. Velocity is acceleration times time,
                                  v = a x t = (4.3 m/s²)(10 s) = 43 m/s