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.

snell's law

refractve index = sin incidence / sin refraction

the answer is the first electromagnet (a). this is because it has more coils around the ferromagnetic material. increasing the number of coils increases the surface area in which the metal interacts with the current in the coil. this enables more atoms in the metal to become aligned hence becoming more magnetized.