a body of mass 200 kg is kept at rest on frictionless surface. it is set into motion when a force of 50 newton is applied on it. calculate the work done by the force in 10 seconds .show that the work done is equal to the change in kinetic energy of the body.
Answers
Answer:
The mass of the body is m=2kg
The force applied on the body F=7N
The coefficient of kinetic friction =0.1
Since the body starts from rest, the initial velocity of body is zero.
Time at which the work is to be determined is t=10s
The acceleration produced in the body by the applied force is given by Newtons second law of motion as:
a
′
=
m
F
=
2
7
=3.5m/s
2
Frictional force is given as:
f=μg=0.1×2×9.8=1.96
The acceleration produced by the frictional force:
a"=−
2
1.96
=−0.98m/s
2
Therefore, the total acceleration of the body:
a
′
+a"=3.5+(−0.98)=2.52m/s
2
The distance traveled by the body is given by the equation of motion:
s=ut+
2
1
at
2
=0+
2
1
×2.52×(10)
2
=126 m
(a) Work done by the applied force,
W
a
=F⋅s=7×126=882 J
(b) Work done by the frictional force,
W
f
=F⋅s=1.96×126=247 J
(c), (d)
From the first equation of motion, final velocity can be calculated as:
v=u+at
=0+2.52×10=25.2m/s
So, the change in kinetic energy is
ΔK=
2
1
mv
2
−
2
1
mu
2
=
2
1
2(v
2
−u
2
)=(25.2)
2
−0
2
=635 J
The distance traveled by the body is given by the equation of motion:
s=ut+
2
1
at
2
=0+
2
1
×2.52×(10)
2
=126 m
Net force =7+(1.96)=5.04 N
Work done by the net force,
W
net
=5.04×126=635 J
Explanation:
its just a verification by an example,for proof use variables in place of values to get relation.
