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Question

A 1.5 kg block at rest on a tabletop is attached to a horizontal spring having a spring constant of 19.6 N/m. The spring is initially unstretched. A constant 20.0 N horizontal force is applied to the object causing the spring to stretch.Determine the speed of the block after it has moved 0.30 m from equilibrium if the surface between the block and the tabletop is frictionless.


A
2.61 m/s
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B
3.61 m/s
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C
7.61 m/s
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D
8.1 m/s
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Solution

The correct option is B 3.61 m/s
This system will exhibits S.H.M with angular frequency ω=km=19.61.5
k= spring constant
m= mass of the body
The string stretched by the maximum A restoring force at maximum stretch= force acting on body
kA=F
A=Fk=2019.6
F=20N
K=19.6N/m
Now if x is displacement from mean position, the velocity is given by:
v=ω=(A2x2)
v=ω=19.61.5(2019.620.32)
=3.523m/s

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