Questions: A box of mass m=6.0 kg (moving in the positive x-direction) has a velocity of 2 m / s at t=0, and has a velocity of 4.0 m / s at time t=2.0 s. From t=0 to t=2.0 s, a force F2 acts on the box in the direction of motion. Find the work done by Fz. Express your answer with the appropriate units. W= Units

Solution Steps

We are given:

• Mass of the box, \( m = 6.0 \, \text{kg} \)
• Initial velocity, \( v_i = 2.0 \, \text{m/s} \)
• Final velocity, \( v_f = 4.0 \, \text{m/s} \)
• Time interval, \( t = 2.0 \, \text{s} \)

The work done by the force \( F_z \) is equal to the change in kinetic energy of the box. The kinetic energy \( K \) is given by: \[ K = \frac{1}{2} m v^2 \]

The initial kinetic energy \( K_i \) is: \[ K_i = \frac{1}{2} \times 6.0 \, \text{kg} \times (2.0 \, \text{m/s})^2 = \frac{1}{2} \times 6.0 \times 4.0 = 12.0 \, \text{J} \]

The final kinetic energy \( K_f \) is: \[ K_f = \frac{1}{2} \times 6.0 \, \text{kg} \times (4.0 \, \text{m/s})^2 = \frac{1}{2} \times 6.0 \times 16.0 = 48.0 \, \text{J} \]

The work done \( W \) by the force \( F_z \) is the change in kinetic energy: \[ W = K_f - K_i = 48.0 \, \text{J} - 12.0 \, \text{J} = 36.0 \, \text{J} \]

Final Answer