A2.3 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?

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 .

the simple machine has three main features. they are:

fulcrum is the center point or pivot point of the machine. effort is input force/ applied force. load is external force/ output.

the keisha 's machine labels are

x : load

y: fulcrum

z: lever

the current is 0.6

voltage ÷ resistance = current

so 12 ÷ 20 = 0.6