Question

A constant force acts on an object of mass 8 kg for a duration of 6 s. It increases the object’s velocity from 3 m s–1 to 15 m s-1. Now, if the force was applied for a duration of 10 s, what would be the final velocity of the object?

A. 13 m/s

B. 35 m/s

C. 23 m/s

D. 28 m/s

Answer 1

Answer:

c. 23 m/s

Explanation:

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Answer 2

answer is option (c) 23m/s

u = 3m/s

v = 15m/s

t = 6s

a = v-u/t

a = 15-3/6

a = 2m/s²

then find force (not compulsory)

f = ma

f = 8×2

f = 16N

new velocity when force acts for time is 10s

v = u+at

v = 3 + 2×10

v = 23m/s

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