A 2.60kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0310m . The spring has force constant 810N/m . The coefficient of kinetic friction between the floor and the block is 0.39 . The block and spring are released from rest and the block slides along the floor.

A-What is the speed of the block when it has moved a distance of 0.0120m from its initial position? (At this point the spring is compressed 0.0190m .)

Express your answer with the appropriate units.

I only got one attempt remaining.

I tried.. (.4323 , .4276 , -.39 , -.62)

0.4323 m/s

Explanation:

We are given;

Mass; m = 2.60kg

Spring Constant; k = 810N/m

coefficient of kinetic friction; μ_k = 0.39

When we apply Newton's second law of motion to the block in the vertical direction, we get;

Σy = N - mg = 0

So, N = mg

Mow,we know that formula for kinetic friction is;

F_k = μ_k•mg

Thus, F_k = 0.39 x 2.6 x 9.8 = 9.9372N

Now, we know that; work = Force x distance.

Since the block when it has moved a distance of 0.0120m from its initial position. Work done by frictional force is;

W_f = - 9.9372 x 0.012 = - 0.119J

The negative sign implies that the frictional force is opposite to the direction of displacement.

Formula for kinetic energy is; K = (1/2)mv²

While elastic potential Energy is given as;

U = (1/2)kx²

Thus,at initial point,

K1 = (1/2) x 2.6 x 0² = 0

U1 = (1/2) x 810 x 0.031² = 0.3892 J

At second point;

K2 = (1/2) x 2.6 x (v_2)² = 1.3(v_2)² J

U2 = (1/2) x 810 x 0.019² = 0.1462J

Work energy theorem gives that;

K1 + U1 = K2 + U2

Thus,

0 + 0.3892 = 1.3(v_2)² + 0.1462

1.3(v_2)² = 0.3892 - 0.1462

(v_2)² = 0.243/1.3

v_2 =√0.1869 = 0.4323 m/s

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