# Energy and Work Done

1. Oct 5, 2005

### jakeowens

Here's one of my questions:

Workers load a 506 kg safe onto an elevator that lifts it 95 m to the twentieth floor of an office building. How much work did the elevator do on the safe?

Now, to solve this, can't i just use the formula W=F(parallel) * delta S. So that would be W=506kg*9.81m/s (since F=ma) *95m= 471086 J

am i correct here?

My other question has me confused

The shot used by male shot-putters has a mass of 7.26 kg. During a throw a shot is initially swun around in a circle reaching a speed of about 3.5 m/s. It is then accelerated, more or less, in a straight line over a distance of about 1.7 m, leaving the hand at roughly 14 m/s.

(a) How much kinetic energy does the shot initally get in the turning phase?
(b) How much kinetic energy does it pickup in the straight-line portion of the launch?
(c) What's the average force exerted on the shot during this latter part of the launch?

Now i get (a) and (b) i just used the equation KE=1/2mv^2, and came up with 44.47J and 711.48J respectively, but I don't quite see how I am supposed to come up with the average force exerted on the shot in the latter part of the launch.

(alot of what i'm looking for here is just confirmation that i am on the right track, i've been missing to many homework problems and it's destroying my grade)

For my last question i attatched a picture, its of a ball rolling down a circularly curved track. The question is what do the speed, acceleration and kinetic energy do as it rolls down the track?

speed-increases
acceleration-increases
and kinetic energy- increases as well. Am i right in this way of thinking?

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2. Oct 5, 2005

### Staff: Mentor

You are correct.

Careful: You calculated the KE at each given speed. That's OK for (a), but not for (b) which asks for the KE added in the straight-line portion.

For (c) consider the work-energy theorem.

2 out of 3 are correct. Since you didn't describe your thinking, just the result of your thinking, you'll have to figure out which one is wrong on your own.

3. Oct 5, 2005

### jakeowens

So for the KE added in the straight line portion, would i just subtract my (a) 44.47J from (b) 711.48J giving me 667.01J?

Or would that be how i get to (c)? the work energy theorem is KE(final)-KE(initial) which would give me 711.48-44.47=667.01J? but my answer has to be in Newtons so how would i get to that?

pertaining to the ball rolling down the circular track- Speed would be increasing because it is rolling down and gravity is speeding it up, acceleration would be decreasing would it not? As the slope of the curve decreases the rate of acceleration would go down. and kinetic energy would also be increasing, because it is related to speed by KE=1/2mv^2.

I just get confused because depending on which point your looking at, the acceleration could be increasing or decreasing. At the top of the curve acceleration would be increasing, but at the bottom it would be decreasing.

4. Oct 5, 2005

### jakeowens

anyone able to help me here?

5. Oct 6, 2005

### mukundpa

You may calculate the acceleration in terms of slop angle $$\theta$$

OO but friction is there !!!

then calculate angular acceleration about instantaneous axis of rotation that is through Point of contact.It is

$$\frac{5g sin \theta }{7R}$$

so with decrease in angle $$\theta$$ the angular acceleration and hence acceleration decreases.

If the acceleration is decreases still velocity increases, only its rate of increase is slowing down.

6. Oct 6, 2005

### Staff: Mentor

Right.

Remember, it's the work-energy theorem. You have the distance.

Right.

The slope of the curve is steepest at the top, getting more horizontal the further down you go. The acceleration decreases as the ball rolls down. At no point does it increase. (Note: I presume all they are interested in is the linear acceleration of the ball, tangential to its path. There is also a radial component of acceleration, due to the ball following a circular path.)