Question

A constant force acts on a object of mass 2kg for 10sec and increases its velocity from 5m/s to 10m/s. Find the magnitude of applied force . If this force was applied for a duration of 15sec, what would be the velocity of the object

Answer 1

M=2kg=
t=10sec
u=5m/sec
v=10m/sec
v=u+at
a=(v-u)/t=(10-5)/10=1/2m/sec²
F=Ma=2×1/2=1kgm/sec²=1N
if this force is applied for 15sec then, velocity is
F=mv/t
v=(1×15)/2=7.5m/sec.

Answer 2

The velocity of object is 12.5 m/s.

SOLUTION:

Let the constant force applied to the object be F

Mass of the object m = 2kg.

The force is applied for an interval of t = 10sec.

There is a change in velocity from 5 m/s to 10 m/s, which means.

Initial velocity u = 5m/s  

And final velocity v = 10m/s.

From Newton’s second law  

F=ma

Here, the acceleration $a=\frac{v-u}{t}=\frac{10-5}{10}=0.5 \mathrm{m} / \mathrm{s}^{2}$

So, the force applied is

$F=2 \times 0.5=1 N$

The applied force becomes 1N.

If the same force is applied for the interval t = 15s.

The value of Velocity will be  

$\begin{array}{c}{v=u+a t} \\ {v=5+(0.5 \times 15)} \\ {=5+7.5} \\ {=12.5 \mathrm{m} / \mathrm{s}}\end{array}$

The velocity of object is 12.5 m/s.