Answer:
You would have to do four times the work W₁ to make the box go the same distance in half the time.
Explanation:
By definition work is:
[tex] W = F*d [/tex] (1)
Where:
F: is the force and d: is the distance
On the other hand, the force is given by:
[tex] F = ma [/tex] (2)
By entering equation (2) into (1):
[tex] W = F*d = m*a*d [/tex] (3)
Also, we can express the acceleration in terms of distance and time by using the following kinematic equation:
[tex] d_{f} = d_{0} + v_{0}t + \frac{1}{2}at^{2} [/tex] (4)
Where:
[tex]d_{f}[/tex]: is the final distance
[tex]d_{0}[/tex]: is the initial distance = 0
[tex]v_{0}[/tex]: is the initial speed = 0 (the box starts at rest)
t: is the time
[tex] d_{f} = \frac{1}{2}at^{2} [/tex]
Solving for a:
[tex] a = \frac{2d_{f}}{t^{2}} [/tex] (5)
Initially we have:
t = T₁, W = W₁, [tex]d_{f}[/tex] = d
[tex] a_{1} = \frac{2d}{T_{1}^{2}} [/tex]
[tex] W_{1} = m*a_{1}*d = m*d*\frac{2d}{T_{1}^{2}} = 2m(\frac{d}{T_{1}})^{2} [/tex]
And when the box goes the same distance in half the time:
[tex] t = \frac{T_{1}}{2} [/tex]
The acceleration is (from equation 5):
[tex]a_{2} = \frac{2d_{f}}{t^{2}} = \frac{2d}{(\frac{T_{1}}{2})^{2}} = \frac{8d}{T_{1}^{2}}[/tex] (6)
Finally, the work W₂ in terms of W₁ is:
[tex]W_{2} = m*a_{2}*d = m*d*\frac{8d}{T_{1}^{2}} = 8m(\frac{d}{T_{1}})^{2} = 4*[2m(\frac{d}{T_{1}})^{2}] = 4W_{1}[/tex]
Therefore, you would have to do four times the work W₁ to make the box go the same distance in half the time.
I hope it helps you!
