07A body In motion comes to rest when force is applied on it. The work done by the force on thebody is
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
Answer:
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