Asked by Anonymous
Two identical 1.60 {\rm kg} masses are pressed against opposite ends of a spring of force constant 1.85 {\rm N}/{\rm cm}, compressing the spring by 25.0 {\rm cm} from its normal length.Find the maximum speed of each mass when it has moved free of the spring on a smooth, horizontal lab table.
Answers
Answered by
Grady
Using conservation of energy. Energy transfers from potential spring to kinetic. 1.85N/cm=185N/m
25cm=.25m
PE=.5*185*(.25^2)=5.78125
PE=KE(final)
KE(final)=1/2(toatl mass)*velocity^2
KE=.5(2*1.6)(v^2)
5.78125=.5*2*1.6*v^2
algebra....v=1.901m/s
25cm=.25m
PE=.5*185*(.25^2)=5.78125
PE=KE(final)
KE(final)=1/2(toatl mass)*velocity^2
KE=.5(2*1.6)(v^2)
5.78125=.5*2*1.6*v^2
algebra....v=1.901m/s
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