Question

A force of 6N acts on a body of mass 1.5kg for 2sec. Assuming the body to be initially at rest, find:
(a) its velocity when the force stops acting.
(b) the distance covered in 5sec after the force starts acting.

Answer 1

Form the formula
F=ma
We will find acceleration produced
6N=1.5x a
a=6/1.5
a=4
By the formula
v=u+at
u=0 (given)
v=4x2=8ms¯¹
B)
By the folmula
S=ut+½ at²
U=0
S=4x5x5/2
S=50m
Do follow and mark as brainliest

Answer 2

It's velocity when the force stops acting is 8m/s.

It's velocity when the force stops acting is 8m/s.The distance covered in 5sec after the force starts acting is 50m.

Given,

Force acting on the body=6N

Mass of the body=1.5kg

Time=2sec.

To find,

the velocity when force stops acting,

the distance covered in 5 sec.

Solution:

Initial velocity, u=0

Final velocity, v.

F=ma

6=1.5xa

a=6/1.5

a=4 m/s^2.

v=u+at

v=0+4x2

v=8m/s.

Time given when force starts acting=5sec

Distance, s.

s=ut+1/2at^2

s=0x5+1/2x4x5x5

s=50m.

It's velocity when the force stops acting is 8m/s.

The distance covered in 5sec after the force starts acting is 50m.

#SPJ2