who knows the answer of this question?

Question

When a 4.00 kg object is hung vertically on a certain light spring that obeys Hooke's law, the spring streches 2.50 cm . If the 4.00 kg object is removed , (a) how far will the spring stretch if a 1.50 kg block is hung on it , and (b) hoe much work must an external agent do to stretch the same spring 4.00 cm from its unstretched position ?

Answer 1

First find the spring constant
F=KX
K=4(9.8)/(2.5*10^-2)
=1568N/m
A)1.5(9.8)=1568(x)
x=0.9375cm.

For extenion of 4cm a force of X will be needed
X=1568(4*10^-2)
F=62.72N

Energy=0.5*force*extension
=0.5*62.72*4/100
=12.544j.

Answer 2

You have Hooke law. Youneed to determine the spring constant (K). Since you have the force and you know the extension (the amt of strech). you can calculate K. Then you can calculate the stretch with a 1.5 kg force. For b), you know K and the amt of stretch so you can calculate the force required. WIth the force and the eqn for Work, you can get the answer. good luck