15 J of work was expended moving an 18 kg block. At what velocity did the block move?

15 J of work was expended moving an 18 kg block. At what velocity did the block move?

Detailed Explanation

We can use the work-energy principle to solve this problem. The work done on the block is 15 J, which is equal to the change in the block's kinetic energy. The kinetic energy of the block is given by:

\(K = (1/2) \times m \times v^2\)

Where m is the mass of the block and v is its velocity.

Substituting the given values, we have:

\(15 J = (1/2) \times 18 kg \times v^2\)

Simplifying, we get:

\(v^2 = 15 J / (9 kg)\)

\(v^2 = 1.67 m^2/s^2\)

Taking the square root, we get:

\(v = sqrt(1.67) m/s\)

\(v ≈ 1.3 m/s\)

Therefore, the block moved at approximately 1.3 m/s. So the answer is 1.3 m/s.

\(K = (1/2) \times m \times v^2\)

Where m is the mass of the block and v is its velocity.

Substituting the given values, we have:

\(15 J = (1/2) \times 18 kg \times v^2\)

Simplifying, we get:

\(v^2 = 15 J / (9 kg)\)

\(v^2 = 1.67 m^2/s^2\)

Taking the square root, we get:

\(v = sqrt(1.67) m/s\)

\(v ≈ 1.3 m/s\)

Therefore, the block moved at approximately 1.3 m/s. So the answer is 1.3 m/s.