3) In the figure, a 5.00-kg block is moving at 5.00 m/s along a horizontal frict

Question

3) In the figure, a 5.00-kg block is moving at 5.00 m/s along a horizontal frictionless surface toward an ideal massless spring that is attached to a wall. After the block collides with the spring, the spring is compressed a maximum distance X, measured in meters. Knowing that the speed of the block when it has moved so that the spring is compressed to only one-half of the maximum distance is equal to 4.3 m/s, what is the value for X?

k=270.33

0.68m

0.78m

0.98m

6.8m

0.068m

Answer: A

Need help figuring pout how to do it

Explanation / Answer

For initial situation

Initial kinetic energy of the block, K1=1/2(mv2)=1/2*(5*52)=62.5 J

Initial potional energy of the spring, P1=0 J

For final situation ,

Final kinetic energy of the block is,k2=0 J

Final potential energy of the spring is, P2=1/2(kx2) J

On appliying the energy conservation theorm,we get

1/2(kx2)=62.5

x=[(2*62.5)/270.33]1/2=0.6799=0.68 m

So option 1st is correct as x=0.68

Related Questions

Navigate

• Browse (All)
• Browse 3
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.