Question #153865

A long spring with a spring constant of 22 N/m is compressed 94 centimetres. a. What is the magnitude of the force exerted on this spring to compress it?

b. The force exerted to compress the spring was actually due to the weight of a block attached to the end of the spring. What is the block’s mass?


Expert's answer

a) The magnitude of the force exerted on this spring to compress it can be found from Hook's law:


F=kxF = kx

where k=22N/mk = 22N/m is the spring constant, x=94cm=0.94mx = 94cm = 0.94m is the compression of the spring. Thus, obtain:


F=22N/m0.94m=20.68NF = 22N/m\cdot 0.94m = 20.68N

b) By definition, the weight of a body is:


P=mgP = mg

where mm is its mass, and g=9.81m/s2g = 9.81m/s^2 is the gravitational acceleration. Since F=PF = P, obtain:


F=mgm=Fgm=20.68N9.81m/s22.1kgF = mg\\ m = \dfrac{F}{g}\\ m = \dfrac{20.68N}{9.81m/s^2} \approx 2.1kg

Answer. a) 20.68 N, b) 2.10 kg.


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