A2.25 kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0340 m . the spring has force constant 855 n/m . the coefficient of kinetic friction between the floor and the block is 0.45 . the block and spring are released from rest and the block slides along the floor. what is the speed of the block when it has moved a distance of 0.0200 m from its initial position? (at this point the spring is compressed 0.0140 m .)

v = 0.434 m/s

Explanation:

let k be the spring force constant. for all the calculations.

the energy stored in the spring is given by:

Es = 1/2×k×(x^2)

    = 1/2×855×(0.0340)^2

    = 0.49419 J

then, the work done by friction is given by:

Wf = f×x = μ×m×g×x = (0.45)×(2.25)×(9.8)×(0.0200) = 0.19845 J

then, the energy stored in the spring at 0.0200 m is:

E = 1/2×k×x^2 = 1/2×(855)×(0.0140)^2 = 0.08379 J

by conservation of energy:

            Es = Wf + E + 1/2×m×v^2

0.49419 J = 0.19845 J + 0.08379 J + 1/2×m×v^2

0.49419 J = 0.19845 J + 0.08379 J + 1/2×(2.25)×v^2

  0.21195 = 1.125×v^2

             v = 0.434 m/s

Therefore, the speed of the block when it has moved 0.0200 from the initial position is 0.434 m/s.

we need answer choices

the current is 0.6

voltage ÷ resistance = current

so 12 ÷ 20 = 0.6