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2 years ago

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A 3.00 kg object is fastened to a light spring, with the intervening cord passing over a pulley. The pulley is frictionless, and

its inertia may be neglected. The object is released from rest when the spring is unstretched. If the object drops 10.0 cm before stopping, find (a) the spring constant of the spring and (b) the speed of the object when it is 5.00 cm below its starting point.

1 answer:

finlep [7]2 years ago

3 0

Answer:

Part a)

k = 588.6 N/m

Part b)

v = 0.7 m/s

Explanation:

As we know that initially block is at rest

now if block is released from rest then it will go down by 10 cm and again comes to rest

so here we have

Part a)

Work done by gravity + work done by spring force = change in kinetic energy

$W_g + W_{spring} = 0 - 0$

$mg(0.10) + \frac{1}{2}k(0^2 - 0.10^2) = 0$

$3(9.81)(0.10) - \frac{1}{2}k(0.10)^2 = 0$

$k = 588.6 N/m$

Part b)

Now when spring is stretch by x = 5 cm then the speed of the block is given as

$mgx' + \frac{1}{2}k(0^2 - x'^2) = \frac{1}{2}mv^2 - 0$

here we have

$x' = 0.05 m$

$3(9.81)(0.05) - \frac{1}{2}(588.6)(0 - 0.05^2) = \frac{1}{2}(3) v^2$

$1.4715 - 0.736 = 1.5 v^2$

$v = 0.7 m/s$

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