a body of mass 2kg is moving in a vertical circular path radius o. 5m . then difference between kinetic engeries at its highest and lowest point s
Answers
Answered by
3
Answer:
ANSWER
Given,
r=1m
m=1kg
g=10m/s
2
In limiting case, for body to revolve in circle,
V
top
=v
1
=
rg
V
bott
=v
2
=
5rg
The difference in kinetic energy,
ΔE=(K.E)
bott
−(K.E)
top
Δ=
2
1
m(v
2
2
−v
1
2
)
Δ=
2
1
m(5rg−rg)=
2
4mrg
ΔE=
2
4×1×1×10
ΔE=20J
The correct option is A.
solution
Explanation:
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Answered by
0
Answer:
Kinetic energy is the energy possessed by a body of mass m due to its velocity v.
KE=
2
1
mv
2
At the highest point velocity, v
A
=
rg
∴KE=
2
1
mv
A
2
At the lowest point velocity, v
B
=
5rg
KE=
2
1
mv
B
2
Difference in KE=
2
1
m(v
B
2
−v
A
2
)=
2
1
m(5rg−rg)
=2mrg=2×1×1×10=20J
Explanation:
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