A 2.60 kg block on a horizontal floor is attached to a horizontal spring that is

Question

A 2.60 kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0340 m . The spring has force constant 815 N/m . The coefficient of kinetic friction between the floor and the block is 0.42 . The block and spring are released from rest and the block slides along the floor.

Part A

What is the speed of the block when it has moved a distance of 0.0100 m from its initial position? (At this point the spring is compressed 0.0240 m .

I tried this using K1+U1+Wother=K2+U2 and got .0938 m/s but it was not correct.

I think i may have messed up the distances in the equations.

If you could show the steps that would be very helpful thank you.

Explanation / Answer

Here ,

let the speed of the block is v m/s

using conservation of energy

0.5 * k * x^2 - 0.5 * k * xf^2 = 0.5 * m * v^2 - uk * m * g * d

0.5 * 815 * (0.0340^2 - 0.0240^2) = 0.5 * 2.6 * v^2 - 0.42 * 2.6 * 9.8 * 0.01

solving for v

v = 0.52 m/s

the speed of the block after travelling 0.01 m is 0.52 m/s

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