Question

A constant force acts on an object of mass 5 kg for a duration of 2s. it increases the object's velocity from 3m/s to 7m/s. Find the magnitude of the applied force. Now, if the force was applied for a duration of 5s, what would be the final velocity of the object?​

Answer 1

Answer:

10 N, 13m/s

Explanation:

m=5kg

u=3m/s

v=7m/s

t=2s

a=(v-u)/t = (7-3)/2 = 4/2 = 2m/s²

F=m×a = 5×2 = 10 N

If t=5s,

v = u+at = 3+2×5 = 3+10 = 13m/s