Asked by Anonymous
When a 4.48 kg mass is hung vertically on a certain light spring that obeys Hooke's law, the spring stretches 2.68 cm.
How much work must an external agent do to stretch the spring 3.96 cm from its unstretched position?
How much work must an external agent do to stretch the spring 3.96 cm from its unstretched position?
Answers
Answered by
bobpursley
from the mass (remember mg is weight), find the spring constant.
Work=1/2 kx^2
Work=1/2 kx^2
Answered by
omar
128.45
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