A moving 2.7 kg block collides with ahorizontal spring whose spring constant is

Question

A moving 2.7 kg block collides with ahorizontal spring whose spring constant is 426N/m. The block compresses the spring a maximumdistance of 15.0 cm from its rest position. Thecoefficient of kinetic friction between the block and thehorizontal surface is 0.14.

A.) What is the work done by the spring in bringing the block torest?

B. )How much mechanical energy is being dissipatedby the force of friction while the block is being brought to restby the spring?

C.) What is the speed of the block when it hitsthe spring?

Explanation / Answer

Energy conservation!!! M= 2.7 k = 426 x = .15 =.14 We are going to use the basic energy conservationequation Kinetic - Wfriction =Potentialspring Kinetic = 1/2mV02 Wfriction = Normal force x x distance ormgd Potentialspring = 1/2kx2 Now plug in our new equations 1/2mV02 - mgd =1/2kx2 Now plug in our values and we find that V0 = 1.99 m/s Potentialspring = 4.79 J Wfriction = .56 J Cheers Buddy Billy Now plug in our new equations 1/2mV02 - mgd =1/2kx2 Now plug in our values and we find that V0 = 1.99 m/s Potentialspring = 4.79 J Wfriction = .56 J Cheers Buddy Billy

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