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A 0.7 kg block attached to a spring with force constant 160 Nm is free to move on a frictionless, horizontal surface. The block is released from rest when the
spring is stretched + 0.15 m as shown in figure below. At the instant the block is released the force vector on the block is.
(Consider Right direction is positive direction and Left direction is negative direction)
- 34.29 N
+ 34.29 N
+ 24 N
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h Oa-241

Respuesta :

Answer:

+ 24 N

Explanation:

the computation is shown below:

Given that

Mass of the block = m = 0.7 kg

Sprint constant = k = 160 N / m

x = 0.15m

Now the force on the block is

F = kx

= (160) (0.15)

= 24 N

As the instant block is released so the acting of the force on the block is positive and it would be in a positive direction i.e. right direction

Therefore the third option is correct

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