### Kinetic Energy and Work Done**Problem Statement:**An accelerating object of mass \( m = 2 \) kg is fired with an initial speed \( v_i = 8 \) m/s at a starting point. If its kinetic energy at its final destination is \( K.E = 268 \) J, what is the resultant work (in joules) done on the object?**Options:**A. 204 B. None C. 704 D. 1004 E. 1204 **Explanation and Solution:**To solve for the resultant work done on the object, we can use the Work-Energy Theorem, which states that the work done on an object is equal to the change in its kinetic energy.The kinetic energy (K.E) of an object is given by:\[ K.E = \frac{1}{2}mv^2 \]1. Calculate the initial kinetic energy (\( K.E_i \)) at the starting point:\[ K.E_i = \frac{1}{2} m v_i^2 \]\[ K.E_i = \frac{1}{2} \times 2 \times (8)^2 \]\[ K.E_i = 1 \times 64 \]\[ K.E_i = 64 \text{ J} \]2. Given the final kinetic energy (\( K.E_f \)) at the final destination is 268 J.3. Calculate the work done (W) on the object:\[ W = K.E_f - K.E_i \]\[ W = 268 \text{ J} - 64 \text{ J} \]\[ W = 204 \text{ J} \]Therefore, the correct answer is option A: 204 J.

Given data: Mass m =2 kg Velocity v=8 m/s2. Final Kinetic energy =268 J

An accelerating object of mass m=2 kg is fired with an initial speed v=8 m/s at a starting point. If its kinetic energy at its final destination is K.E=268 J. What is the resultant work (in joules) done on the object? A. 204 O. B. None C. 704 D. 1004 E. 1204

An accelerating object of mass m=2 kg is fired with an initial speed v=8 m/s at a starting point. If its kinetic energy at its final destination is K.E=268 J. What is the resultant work (in joules) done on the object? A. 204 O. B. None C. 704 D. 1004 E. 1204

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Instantaneous Power

Power is defined in physics as the amount of energy transmitted or transformed per unit time. The watt is the unit of electricity in the International System of Units, equivalent to one joule per second.

Question

### Kinetic Energy and Work Done

**Problem Statement:**

An accelerating object of mass \( m = 2 \) kg is fired with an initial speed \( v_i = 8 \) m/s at a starting point. If its kinetic energy at its final destination is \( K.E = 268 \) J, what is the resultant work (in joules) done on the object?

**Options:**

A. 204
B. None
C. 704
D. 1004
E. 1204

**Explanation and Solution:**

To solve for the resultant work done on the object, we can use the Work-Energy Theorem, which states that the work done on an object is equal to the change in its kinetic energy.

The kinetic energy (K.E) of an object is given by:
\[ K.E = \frac{1}{2}mv^2 \]

1. Calculate the initial kinetic energy (\( K.E_i \)) at the starting point:
\[ K.E_i = \frac{1}{2} m v_i^2 \]
\[ K.E_i = \frac{1}{2} \times 2 \times (8)^2 \]
\[ K.E_i = 1 \times 64 \]
\[ K.E_i = 64 \text{ J} \]

2. Given the final kinetic energy (\( K.E_f \)) at the final destination is 268 J.

3. Calculate the work done (W) on the object:
\[ W = K.E_f - K.E_i \]
\[ W = 268 \text{ J} - 64 \text{ J} \]
\[ W = 204 \text{ J} \]

Therefore, the correct answer is option A: 204 J.

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