Question

mechanical energy is defined as: * 1 point
the energy required to move a 1 kg object 1 m
the sum of kinetic and gravitational potential energy
the change in kinetic energy of an object
the change in gravitational potential energy of an object
the difference between kinetic and gravitational potential energy

an object is travelling at a final speed of 12 m/s. if the process was only 68% efficient, what was the initial speed of the object? * 1 point
10 m/s
11 m/s
12 m/s
15 m/s
16 m/s

Explanation:

Step1: Recall the formula for kinetic energy

The formula for kinetic energy is \(K = \frac{1}{2}mv^{2}\). Let the initial speed be \(v_{i}\) and the final speed be \(v_{f}\). The efficiency \(\eta\) is given by \(\eta=\frac{K_{f}}{K_{i}}\), where \(K_{f}=\frac{1}{2}mv_{f}^{2}\) and \(K_{i}=\frac{1}{2}mv_{i}^{2}\).

Step2: Substitute the formula for kinetic energy into the efficiency formula

Since \(\eta=\frac{K_{f}}{K_{i}}\), and \(K_{f}=\frac{1}{2}mv_{f}^{2}\), \(K_{i}=\frac{1}{2}mv_{i}^{2}\), we can cancel out \(\frac{1}{2}m\) (because mass \(m\) is non - zero). So \(\eta=\frac{v_{f}^{2}}{v_{i}^{2}}\).

Step3: Solve for \(v_{i}\)

We know that \(\eta = 0.68\) and \(v_{f}=12\space m/s\). From \(\eta=\frac{v_{f}^{2}}{v_{i}^{2}}\), we can rewrite it as \(v_{i}^{2}=\frac{v_{f}^{2}}{\eta}\). Then \(v_{i}=\sqrt{\frac{v_{f}^{2}}{\eta}}\).

Substitute \(v_{f} = 12\space m/s\) and \(\eta=0.68\) into the formula:

\(v_{i}=\sqrt{\frac{12^{2}}{0.68}}=\sqrt{\frac{144}{0.68}}\approx\sqrt{211.76}\approx14.55\approx15\space m/s\)

Answer:

15 m/s