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A 2.3 kg block is pulled across a friction-free horizontal surface, stretching a spring that has a spring constant of 18 N/m. At the moment it is released, the block accelerates at 0.27 m/s2.
What is the net force that the spring exerts on the block?
N
How far was the spring stretched at the moment it was released?

Respuesta :

AL2006

To find the force on the block, use Newton #1:  F = m A

Force = (2.3 kg) x (0.27 m/s²)

Force = (2.3 x 0.27) kg-m/s²

Force = 0.621 Newton

The spring stretches 1 meter when to pull with 18 N of force.

To pull with only 0.621 N of force, it only has to stretch (0.621/18) of a meter.  That's 0.0345 meter, or 3.45 cm .

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