When a 4.60-kg object is hung vertically on a certain light spring that obeys Hooke's law, the spring stretches 2.30 cm. (a) If
the 4.60-kg object is removed, how far will the spring stretch if a 1.50-kg block is hung on it? cm (b) How much work must an external agent do to stretch the same spring 4.00 cm from its unstretched position?
1 answer:
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Explanation:
direction of electric field is same as that of force experienced by the test charge
