When a 2.90-kg object is hungvertically on a certain light spring described by H
ID: 1679147 • Letter: W
Question
When a 2.90-kg object is hungvertically on a certain light spring described by Hooke's law, thespring stretches 2.76 cm. (a) What is the force constant of the spring?N/m
(b) If the 2.90-kg object isremoved, how far will the spring stretch if a 1.45-kg block is hung on it?
cm
(c) How much work must an external agent do to stretch the samespring 7.20 cm from itsunstretched position?
J (a) What is the force constant of the spring?
N/m
(b) If the 2.90-kg object isremoved, how far will the spring stretch if a 1.45-kg block is hung on it?
cm
(c) How much work must an external agent do to stretch the samespring 7.20 cm from itsunstretched position?
J
Explanation / Answer
a) since considering the only a one dimension of the system, we don'tneed to think about breaking into vector components. force contact of the spring = k F = -kx F= mg = 2.9 kg * 9.81 m/s^2 = 28.4 N x=.0276 m (converted cm to m) -k=F/x ; k=-1030.8 N/m ( the minus sign on thevalue indicates that the direction of the spring force isalways opposite the direction of the displacement of the spring'sfree end.) b) We can use the same formula to find x this time. F now becomes 1.45kg * 9.81 m/s^2 = 14.2 N and by using the k we found from the question a), we can findx. F=-kx ; x=F/-k ; 14.2N/-1030.8 N/m = -.0138 m c) Work done by Spring force = 1/2kx^2(initial)-1/2kx^2(final) The answer come out to be 2.67 J I'm pretty sure the number are correct, but I forgot about puttingcorrect sign for the spring force, please check into that.Sorry!
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