You do a certain amount of work on an object initially at rest, and all the work goes into increasing the object's speed. If you do work W, suppose the object's final speed is v. What will be the object's final speed if you do twice as much work?
1. 4 v
2. 2 v
3. Still v
4.v/�ã2
5. v*�ã2
Please quit making the duplicate posts
To determine the final speed of an object when you do twice as much work, 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.
Let's assume the initial speed of the object is zero, as it is initially at rest.
1. According to the work-energy theorem, the work done on the object initially is equal to the change in its kinetic energy, which can be written as:
W = ΔKE
2. If you do twice as much work (2W), the change in kinetic energy will also be doubled:
2W = 2(ΔKE)
3. Since all the work goes into increasing the object's speed, the change in kinetic energy is equal to the final kinetic energy:
ΔKE = KE_f
4. Rearranging the equation, we have:
2W = 2(KE_f)
5. Dividing both sides of the equation by 2, we get:
W = KE_f
From this equation, we can see that the final kinetic energy (and, therefore, the final speed) of the object will remain the same regardless of how much work you do. Therefore, the correct answer is option 3: Still v.