A 0.60-kg block initially at rest on a frictionless, horizontal surface is acted upon by a force of 7.0 N for a distance of 2.0 m. How much farther would the force have to act for the block to have 57 J of kinetic energy?

Answer:

Explanation:

o have 57 J of kinetic energy?

We can use the work-energy principle to solve this problem. The work-energy principle states that the work done on an object is equal to the change in its kinetic energy. The work done by a constant force on an object is given by the product of the force and the distance over which the force is applied.

Let's first find the initial kinetic energy of the block, which is zero because the block is initially at rest. Then we can find the work done by the force:

W = Fd = (7.0 N)(2.0 m) = 14 J

The work done by the force is 14 J. We want to find the additional distance the force would have to act to give the block a total kinetic energy of 57 J. Let x be the additional distance:

Work done by force over x distance:

W = Fd = (7.0 N)(x) = 7x J

The total work done on the block is the sum of the work done by the force and the change in kinetic energy:

W_total = W + ΔK

where ΔK is the change in kinetic energy.

At the final position, the block has 57 J of kinetic energy, so:

W_total = 57 J

We can now solve for the additional distance x:

W_total = W + ΔK

57 J = 14 J + (1/2)mv_f^2

where v_f is the final velocity of the block.

Since the block starts from rest, the final velocity is given by:

v_f^2 = 2ΔK / m

v_f^2 = 2(57 J) / 0.60 kg = 95 m^2/s^2

v_f = sqrt(95) = 9.746 m/s

Now we can solve for x:

57 J = 14 J + (1/2)(0.60 kg)(9.746 m/s)^2 - (1/2)(0.60 kg)(0 m/s)^2

57 J = 14 J + 27.8 J + 0

57 J - 14 J - 27.8 J = 7x J

15.2 J = 7x J

x = 2.17 m

Therefore, the force would have to act for an additional distance of 2.17 m for the block to have 57 J of kinetic energy.

What other transformation is used to create this pattern