Question

a force acts for 0.1s on a body of mass 1.2kg initially at rest.the force then ceases to act and the body moves through 2m in the next one second. find the magnitude of force.

Answer 1

t=0.1s
m=1.2kg
u=0 m/s
then, acc.=(2 /s)sq.
as the body moved/accelerated 2m in 1sec
f= m x a
i.e.=1.2 x 2= 2.4N
hence force exerted by the body is 2.4N

Answer 2

The magnitude of the force that acted on the body is equal to 24 N.

Step-by-step Explanation

Given:

• The force acting on the body is 0.1s.
• The mass of the body is 1.2 kg.
• The body is initially at rest.
• After the force acting ceases (or stops), the body moves through a distance of 2 m in the next one second.

To be found: To find the magnitude of the force that acted on the body.

Solution:

When the force ceases acting on the body, the force, F = 0

When force is zero, acceleration also becomes zero (a = 0)

Since the acceleration is equal to zero, the velocity of the body is constant.

It is given that the body moves a distance of 2 m in the next 1 second when force ceases acting.

$\implies$ Distance travelled, s = 2m and Time taken (when force is not acting), $t_{1}$ = 1s

Using the speed formula, we have

Speed, v = Distance(s) / Time(t)

$\implies v=\frac{2}{1}$

$\implies v = 2$ m/s   ------(1)

Now, consider the case when force is acting.

Here, the mass of the body, m = 1.2 kg; the time taken (for force acting), $t_{2}$ = 0.1s.

Also, given that the body is initially at rest, which means that the initial velocity, u = 0.

From (1), we have the final velocity, v = 2 m/s

Hence, using the formula, $a=\frac{v-u}{t}$ here, we get the acceleration of the body as,

$a=\frac{v-u}{t_{2} } \\\implies a=\frac{2-0}{0.1}$

$\implies a=20$ $m/s^{2}$

Now, calculating the magnitude of force, we get

Force, $F=m \times a$

$\implies F=1.2 \times 20\\\implies F=24 N$

Therefore, the magnitude of the given force that acted on the body is 24 N.

#SPJ3