A block of mass 2.5 kg starting from rest falls a verticaldistance of 75 cm befo

Question

A block of mass 2.5 kg starting from rest falls a verticaldistance of 75 cm before striking a vertical coiled spring. As aresult of this contact the spring is compressed a distance X fromits original length when the block comes momentarily to rest. TheSpring constant for the Spring is 4000N/m

A. What is the Distance X that the Spring is compressed?
B. After the block is momentarily at rest, the spring will returnto its equilibrium position and eject the block back into the air.What is the Speed of the Block when it is a distance of 50 cm abovethe end of the spring?

Please show all work and formulas used to get the answer. Thankyou

Explanation / Answer

a) PEi=PEf, so mgh=1/2kx2 thenx=((2mgh)/k)=((2(2.5)(9.8)(7x10-2m)/4000)=9.58cm b) Now we have PEi=PEf+KEf, so1/2kx2=mgh+1/2mv2 thenv2=2(1/2kx2-mgh)/m=2(1/2(4000)(9.58x10-2)2-(9.8)(2.5)(50x10-2))/2.5kg=4.88m2/s2then v=2.21m/s

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