RRB Group D 24-Sep-2018 Shift 1 Solved Paper

Question : 98 of 100

Marks: +1, -0

When a constant force acts on an object with a mass of

8 \ { kg}

for a duration of

2 \ { s}

, it increases the object's velocity from

4 \ { ms}^{-1}

9 \ { ms}^{-1} .

What is the magnitude of the applied force?

Solution:

Let the Force be F, mass

=8 \ { kg}

{\time}({t})=2 {s}

Initial velocity

({u})=4 {m} \/ {s}

Final velocity

({v})=9 {m} \/ {s}

So, let the acceleration be a
So,

a=(v-u) / t=(9-4) / 2

=2.5 {m} \/ {s}^{2}

{F}={m}×{a}=8 \×2.5

i.e

20 { N}

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