Explain law of conservation of energy
on the basis of oscillating spring on
the friction less table. Spring constant
of the spring is 500 N/m as shown in
the figure. A body of mass 1kg is
moving with the velocity of 20 m/s. Find the compression that occur in
the spring.
Answers
Answered by
0
Answer:
500effective to become the decimal expansion of the hardships underwent by sachin
Answered by
0
Answer:
(a) At Equilibrium,
Kx−7=0
Kx=7
x=7/K=10/100=0.1m
(b) PE+KE=
2
1
mv
2
+
2
1
Kx
2
=
2
1
(1)(2)
2
+
2
1
(100)(0.1)
2
=
2
5
J
(c) we know that,
in SHM T=2π
k
m
T=2π
100
1
=
5
π
=0.628s
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