A block with mass M attached to a horizontal spring with force constant k is moving with simple harmonic motion having amplitude
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Answer:
<em>The rubber band will be stretched 0.02 m.</em>
<em>The work done in stretching is 0.11 J.</em>
Explanation:
Force 1 = 44 N
extension of rubber band = 0.080 m
Force 2 = 11 N
extension = ?
According to Hooke's Law, force applied is proportional to the extension provided elastic limit is not extended.
F = ke
where k = constant of elasticity
e = extension of the material
F = force applied.
For the first case,
44 = 0.080K
K = 44/0.080 = 550 N/m
For the second situation involving the same rubber band
Force = 11 N
e = 550 N/m
11 = 550e
extension e = 11/550 = <em>0.02 m</em>
<em>The work done to stretch the rubber band this far is equal to the potential energy stored within the rubber due to the stretch</em>. This is in line with energy conservation.
potential energy stored =
==> = <em>0.11 J</em>
Explanation:
Given
initial speed(u)=3 m/s
mass of each ball is m
Let the cue ball is moving in x direction initially
In elastic collision Energy and momentum is conserved
Let u be the initial velocity and be the final velocity of 8 ball and cue ball respectively
The angle after which cue ball is deflected is given by
Conserving momentum in x direction
Along Y axis
substitute the value of
we get