Question

A force of 62 N acts on a 35kg object for 10.0s. What is the object change in velocity?

Answer 1

A force of 62 N acting on a 35 kg object applies an acceleration a of

62 N = (35 kg) a

a = (62 N) / (35 kg) ≈ 1.8 m/s²

If v₀ is the object's initial velocity, then after 10.0 s it attains a final velocity v of

v = v₀ + a t

v = v₀ + (1.8 m/s²) (10.0 s)

v = v₀ + 18 m/s

so that the change in velocity is

∆v = v - v₀ = 18 m/s

Answer 2

not sure but 17.714 smthng like that

Step-by-step explanation:

first find a=62/35 from f=m*a then v=a*t

(62/35)*10