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Physics - SHM Question with Solution | TestHub

PhysicsSHMEnergyMedium2 minQB

PhysicsMediumnumerical

A block of mass $m = 2 kg$ is connected to a spring of force constant $k = 50 N / m$ . Initially the block is at rest and the spring has natural length. A constant force $F = 60 N$ is applied horizontally towards right, the maximum speed of the block (in $m / s$ ) will be (neglect friction):

Answer:

6.00

Solution:

Ans (6.00)

For maximum speed, $\frac{d v}{d t} = 0 \Rightarrow a = 0$

If at maximum speed, elongation is x, then $F = k x \Rightarrow x = \frac{F}{k}$

Now, by work-energy theorem: $- \frac{1}{2} k x^{2} + F x = \frac{1}{2} m v^{2}$

$∴ v = \frac{F}{mk} = 6$ m/s

Alternate: $v_{ma x} = A ω = \frac{F}{k} \frac{k}{m} = \frac{F}{km}$

Stream:JEESubject:PhysicsTopic:SHMSubtopic:Energy

⏱ 2mℹ️ Source: QB

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