Answer to Question #330661 in Physics for Jane

Question #330661

The work done on a stretching a spring it to 3.50 cm from its equilibrium length is 14.0 3. (a) How much force is required to stretch it at that





distance? (b) If a block of mass 2.75 kg moving at 4.8 m/s hits and runs into the same spring at its unstretched length, what is maximum compression of the spring?

1
Expert's answer
2022-04-21T13:02:20-0400

Given:

x=0.035mx=0.035\:\rm m

W=14.0JW=14.0\:\rm J

(a) the required force

F=2Wx=214.00.035=800NF=\frac{2W}{x}=\frac{2*14.0}{0.035}=800\:\rm N

(b) the stifness of a spring

k=Fx=8000.035=2.29104N/mk=\frac{F}{x}=\frac{800}{0.035}=2.29*10^4\:\rm N/m

hence, the maximum compression of a spring

xmax=vm/k=4.82.75/2.29104=0.053m=5.3cmx_{\max}=v\sqrt{m/k}\\ =4.8\sqrt{2.75/2.29*10^4}=0.053\:\rm m=5.3\: cm


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