To determine the final speed of the object after 150 J of work is done on it, we can apply the work-energy principle. This principle states that the work done on an object is equal to the change in its kinetic energy:

$W = \Delta KE$

Step-by-Step Solution:

1. Initial Kinetic Energy:

   The initial kinetic energy $KE_i$ is given by:

   $KE_i = \frac{1}{2}mv_i^2$

   where:

   • $m = 1.00 \, \text{kg}$ is the mass of the object,
   • $v_i = 10.0 \, \text{m/s}$ is the initial velocity.

   Substituting the values:

   $KE_i = \frac{1}{2} \times 1.00 \times (10.0)^2 = 50.0 \, \text{J}$

2. Work Done:

   The work done on the object is 150 J. Since work increases the object's kinetic energy, the final kinetic energy $KE_f$ will be:

   $KE_f = KE_i + W = 50.0 \, \text{J} + 150.0 \, \text{J} = 200.0 \, \text{J}$

3. Final Velocity:

   The final kinetic energy is related to the final velocity by:

   $KE_f = \frac{1}{2}mv_f^2$

   Solving for the final velocity $v_f$:

   $v_f = \sqrt{\frac{2 KE_f}{m}}$

   Substituting the known values:

   $v_f = \sqrt{\frac{2 \times 200.0}{1.00}} = \sqrt{400.0} = 20.0 \, \text{m/s}$

Final Answer:

The object will be moving at 20.0 m/s after 150 J of work is done on it.

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Do you want further details on any step or have additional questions?

Here are 5 related questions to think about:

1. How would the result change if the object had more mass?
2. What happens if the work done was negative (e.g., due to friction)?
3. Can we calculate the time taken for the work to be done?
4. How does the direction of the force affect the work done on the object?
5. What if the object started at rest—how would the result differ?

Tip: Always remember that work done on an object can either increase or decrease its kinetic energy, depending on whether the force is applied in the same or opposite direction as the motion.