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/(sqrt2)
5. v*(sqrt2)
To determine the object's final speed when you do twice as much work, we need to understand the relationship between work done on an object and its final speed.
The work-energy principle states that the work done on an object is equal to the change in its kinetic energy. Mathematically, we can write it as:
W = ΔKE
where W is the work done on the object, and ΔKE represents the change in the object's kinetic energy.
The kinetic energy (KE) of an object is given by:
KE = 1/2 * m * v^2
where m is the mass of the object, and v is its velocity or speed.
If we initially do work W on an object at rest, its initial kinetic energy is zero. So, the work done is entirely converted into the change in kinetic energy (ΔKE):
W = ΔKE
When the object starts with an initial velocity of zero and the work goes into increasing its speed to v, we have:
W = KE_final - KE_initial
W = (1/2 * m * v^2) - (1/2 * m * 0^2)
W = (1/2 * m * v^2) - 0
W = 1/2 * m * v^2
To determine the object's final speed when we do twice as much work (2W), we can set up an equation based on the work-energy principle:
2W = ΔKE
Substituting the expression for work (W) and using the formula for kinetic energy (KE), we can write:
2(1/2 * m * v^2) = 1/2 * m * v_f^2
Simplifying the equation:
2 * 1/2 * m * v^2 = 1/2 * m * v_f^2
m * v^2 = 1/2 * m * v_f^2
Cancelling out the mass (m) and simplifying further:
v^2 = 1/2 * v_f^2
2v^2 = v_f^2
Taking the square root of both sides:
√(2v^2) = √(v_f^2)
√2 * v = v_f
Therefore, the object's final speed when we do twice as much work will be v multiplied by the square root of 2, which means option 5 is the correct answer: v * √2.