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Elodia [21]
3 years ago
10

One end of a horizontal spring with force constant 130.0 Ni'm is attached to a vertical wall. A 4.00-kg block sitting on the flo

or is placed against the spring. The coefficient of kinetic friction between the block and the floor is uk = 0.400. You apply a constant force F to the block. F has magnitude F = 82.0 N and is directed toward the wall. At the instant that the spring is com-pressed 80.0 cm, what arc (a) the speed of the block, and (b) the magnitude and direction of the block's acceleration?
Physics
1 answer:
tekilochka [14]3 years ago
4 0

Answer:

Explanation:

When the spring is compressed by .80 m , restoring force by spring on block

= 130 x .80

= 104 N , acting away from wall

External force = 82 N , acting towards wall

Force of friction acting towards wall = μmg

= .4 x 4 x 9.8

= 15.68 N

Net force away from wall

= 104 -15.68 - 82

= 6.32 N

Acceleration

= 6.32 / 4

= 1.58 m / s²

It will be away from wall

Energy released by compressed spring = 1/2 k x²

= .5 x 130 x .8²

= 41.6 J

Energy lost in friction

= μmg x  .8

= .4 x 4 x 9.8 x .8

= 12.544 J

Energy available to block

= 41.6 - 12.544 J

= 29 J

Kinetic energy of block = 29

1/2 x 4 x v² = 29

v = 3.8 m / s

This will b speed of block as soon as spring relaxes. (x = 0 )

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Answer:

Explanation: Ok, first caracterize the two vectors that we know.

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<span>Thank you for posting your question. I hope you found what you were after. Please feel free to ask me more.</span>


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