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shepuryov [24]
3 years ago
15

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.
Physics
1 answer:
finlep [7]3 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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Explanation:

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First, let us see if Net force = MA. To do that, we need to determine whether the object is at equilibrium horizontally. For an object to be at equilibrium, it either needs to be moving at a constant velocity or not moving at all. Also, if an object is at equilibrium, there will not be any acceleration. But we know that there IS acceleration horizontally, so it cannot be in equilibrium. If it is not in equilibrium, we can use the formula ∑F= ma.

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μk is the coefficient, which the problem includes- it is 0.613.

N is the normal force, which we have to find.

*To find the normal force, we have to determine if the object is at equilibrium VERTICALLY. Since it has no acceleration vertically (it's not moving up/down), it is at equilibrium. Now, when an object is at equilibrium in one direction, it means that all the forces in that direction are equal. What are our vertical forces? Weight (mg) and Normal force (N). So it means that the Normal force is equal to the Weight.

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Normal force = 392 N

Now, plug it back into the formula (μkN): (0.613)(392) = 240.296 N

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Now that we know the friction, we can find the horizontal net force. Just subtract the friction force, 240.296 from the applied force, 1055 N

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Now that we know the net force, plug in the numbers for the formula

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