a horizontal spring is stretched from its equilibrium position when a force of 0.75kg acts on it. then a force of 1.5 kg is attached to the end of the end of the spring and pulled 4.5 cm along a frictionless table from the equilibrium position. the body is then released and performs a simple harmonic motion calculate (1) the force constant of the spring (2)force exerted by the spring on the 1.5kg body before it was released?
working
f=ke
0.75+1.5=2.25kg
2.25kg*10 =22.5n
k =f/e= 22.5n/0.045m= 500nm
but I checked the answer from the answer page my answers were wrong please help oo.
0 answers