Question

How do you calculate work needed when only given mass (in kg) and m/s? Can someone please do an example problem?

Answer 1

I'm not sure if I understand... are you asking how to find out how much work it takes to accelerate an object of mass \(m\) from rest to a certain velocity \(v\)?

Answer 2

For example-- how much work is needed to receive a 2.3 kg basketball that is traveling at 60 m/s?

Answer 3

One way to define work is as a change in kinetic energy. Kinetic energy \(E_k\) is given generically for mass \(m\) and velocity \(v\) by \(\boxed{E_k=\frac{1}{2}mv^2}\). The change in energy \(\Delta E\) (which is identical tow work \(W\)) is just the final kinetic energy minus the initial (as with any "change" variable). The final kinetic energy, when it is caught, is zero, and the initial is \(E_k=\frac{1}{2}mv^2=\frac{1}{2}\left(2.3 \text{ kg}\right)\left(60 \frac{\text m}{\text s}\right)^2=4140 \text{ J}\) (energy is measured in Joules \(\text J\)). Thus, the change in kinetic energy, or work, is \(\boxed{\Delta E = W = -4140 \text{ J}}\). The work is negative since the ball is going from a state of high energy to low energy EVEN THOUGH it takes effort by you to do so. That's one thing about work that can be hard to understand at first (really because it was given a lousy name). Work is something you transfer, NOT something you do (what you do is your force).

Answer 4

Okay, thanks!! So the work for a problem like this will always be negative?

Answer 5

Yes. If the object is going from high energy to low energy, and the mediator in question (which here is us) is causing the object to go from high energy to low energy, then the work imparted by the mediator (us) is negative.