You do a certain amount of work on an object initially at rest, and all the work goes into increasing the object�fs speed. If you do work W, suppose the object�fs final speed is v. What will be the object�fs final speed if you do twice as much work?
1. 4 v
2. 2 v
3. Still v
4.v/�ã2
5. v*�ã2
Work done is proportional to V^2.
If work doubles, the V must increase by a factor sqrt2
To find the answer, we need to understand the relationship between work done on an object and the change in its kinetic energy.
The work-energy principle states that the work done on an object is equal to the change in its kinetic energy. Mathematically, this can be written as:
Work (W) = Change in kinetic energy (ΔKE)
Since in this case, all the work goes into increasing the object's speed, we can say that the change in kinetic energy is equal to the final kinetic energy (KEf) minus the initial kinetic energy (KEi):
ΔKE = KEf - KEi
Now, let's look at the relationship between kinetic energy and speed. The kinetic energy of an object is given by the equation:
KE = (1/2)mv^2
Where m is the mass of the object and v is its speed.
Since we are only concerned with the change in speed, we can rewrite the equation as:
ΔKE = (1/2)m(vf^2 - vi^2)
Now, substituting this equation back into the work-energy principle equation, we get:
W = ΔKE = (1/2)m(vf^2 - vi^2)
From this equation, we can see that the final speed (vf) is proportional to the square root of the work done (W). Mathematically, we can write:
vf = √(2W / m) [Equation 1]
Now, if we do twice as much work, the new work done will be 2W. Substituting this value into Equation 1, we get:
vf' = √(2(2W) / m)
= √(4W / m)
= √4 * √(W / m)
= 2√(W / m)
Therefore, the object's final speed if you do twice as much work will be 2 times the square root of the original final speed. In other words, the correct answer is:
Option 5: v * √2
Note: None of the provided options match the correct answer exactly.