If a body having mass 40kg started moving initially with rest and it takes a velocity of 20m/sec in time 4 seconds. Find the value of force.
Answers
Step-by-step explanation:
Given:
The mass of body, m=0.40kg
Constant initial speed of the body is, u=10m/s
Force that acts on the body is, F=8.0N
(i)
At t= –5 s
The force starts acting on the body from t=0 s.
So, the acceleration of the body during this time was zero and it moves at a constant speed.
Position of the body is given by:
x=v×t
x=10×(−5)
=−50 m
(ii)
At t=25 s
Since the force acts in the opposite direction of motion of particle. So, the acceleration of the body due to the force acting on it is:
a=
M
F
=
0.40
−8.0
=−20ms
−2
The position is given by the second equation of motion:
s
′
=ut
′
+
2
1
a"t
2
=10×25+
2
1
×(−20)×25
2
=250−6250=−6000m
(iii)
At t=100 s
For first 30 sec, it moves under retardation of force and after that the body moves at constant speed.
a=−20ms
−2
u=10m/s
The distance covered in 30 s:
s
1
=ut+
2
1
a"t
2
=10×30+
2
1
×(−20)×30
2
=300−9000=−8700m
Now, the speed after 30 sec is:
v=u+at
v=10−(20×30)
=−590 m/s
Distance covered in remaining 70 sec:
x =−590×70
= −41300 m
So, total distance covered in 100 s will be:
X=−8700−41300
X=−50000 m=−50 km