Answer :
The final speed of the object is approximately 11.4 m/s.
To determine the final speed of the object, we can use the Work-Energy Theorem. This theorem states that the work done on an object is equal to its change in kinetic energy. The formula for kinetic energy (KE) is:
[tex]KE = \frac{1}{2}mv^2[/tex]
Given:
- Mass (m) = 8.4 kg
- Initial speed ([tex]v_{i}[/tex]) = 9.6 m/s
- Work done (W) = 163 N·m
First, we calculate the initial kinetic energy ([tex]KE_{i}[/tex]) of the object:
[tex]KE_i = \frac{1}{2} m v_i^2[/tex]
Substituting the given values:
[tex]KE_i = \frac{1}{2} \times 8.4 \text{ kg} \times (9.6 \text{ m/s})^2[/tex]
[tex]KE_i = \frac{1}{2} \times 8.4 \times 92.16[/tex]
[tex]KE_i = 4.2 \times 92.16[/tex]
[tex]KE_i = 387.07 \text{ J}[/tex]
Next, the work-energy theorem tells us that the net work done (W) on the object results in a change in kinetic energy, so the final kinetic energy ([tex]KE_{f}[/tex]) is:
[tex]KE_f = KE_i + W[/tex]
[tex]KE_f = 387.07 \text{ J} + 163 \text{ J}[/tex]
[tex]KE_f = 550.07 \text{ J}[/tex]
Now, we use the final kinetic energy to find the final speed ([tex]v_{f}[/tex]):
[tex]KE_f = \frac{1}{2} m v_f^2[/tex]
Solving for [tex]v_f[/tex]:
[tex]550.07 = \frac{1}{2} \times 8.4 \times v_f^2[/tex]
[tex]550.07 = 4.2 \times v_f^2[/tex]
[tex]v_f^2 = \frac{550.07}{4.2}[/tex]
[tex]v_f^2 = 130.97[/tex]
[tex]v_f = \sqrt{130.97}[/tex]
[tex]v_f \approx 11.4 \text{ m/s}[/tex]
