Solution :
Given :
Mass attached to the spring = 4 kg
Mass dropped = 6 kg
Force constant = 100 N/m
Initial amplitude = 2 m
Therefore,
a).
= 10 m/s
Final velocity, v at equilibrium position, v = 5 m/s
Now,
A' = amplitude = 1.4142 m
b).
m' = 2m
Hence,
c).
Therefore, factor
Thus, the energy will change half times as the result of the collision.
Answer:
Explanation:
Work
Other units Foot-pound, Erg
In SI base units 1 kg⋅m2⋅s−2
Derivations from other quantities W = F ⋅ s W = τ θ
Dimension M L2 T−2
Idk if this is what u are looking for but i hope this help.:)
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I'm not sure what is nmeant by "p.d. remains different" .
Answer:
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Explanation: