# Physics Exam Revision Help - Exam in 5 hours

1. Jun 4, 2014

### ichigo

Your help would be greatly appreciated

1. If an object is dropped from a cliff, how many seconds will it take to have a speed of 80m/s?

would gravity be taken into account? This question seems weird because you usually get given 2 variables to get the third, unless I am missing something...

2. What force is required to lift a 20kg weight off the ground?

gravity is 10 (for our exam, 10 is acceptable on earth), 20kg mass. F=ma, but there is no acceleration given, so I don't know what to do

3. A man exerts a 300N upwards force on a 20kg weight that was resting on the ground

a) draw a free body diagram of the forces on the weight
I don't know what a free body diagram is

b) What is the net force on the weight?

I don't know how to calculate net force

Sorry if these sound like easy questions, but this is my first year of studying physics, and I don't get it 100%

Last edited: Jun 4, 2014
2. Jun 4, 2014

### Nathanael

(I assume that) The question assumes that the gravitational acceleration is $9.8\frac{m}{s^2}$ (or $10\frac{m}{s^2}$) so you actually have 3 pieces of information (final velocity = $80\frac{m}{s}$ ...... initial velocity = $0\frac{m}{s}$ ...... acceleration = $9.8\frac{m}{s^2}$ or $10\frac{m}{s^2}$) and you want to figure out the fourth piece of information (time)
Do you know how to do this?

Let's restate the problem:
What force is required to hold a 20kg weight above the ground?

Now can you solve the problem?

Net force depends on the situation (It's the sum of all the forces.)

In this situation what forces act on the weight?
Well I can tell you one force is 300N Upwards (remember! Direction IS IMPORTANT) because that's what the problem says! :)

What other force(s) would be acting on the object? (And in what direction?)

Last edited: Jun 4, 2014
3. Jun 4, 2014

### ichigo

Would the answer to question 2 be 200N?

If the weight of the object is 200N (20x10), wouldn't the force required to lift it be equal and opposite?

Also, for question 3, would the other force be the weight of the object? So 200N, + 300N, is the net force 500N?

4. Jun 4, 2014

### Nathanael

Correct. Equal in magnitude, opposite in direction.

But remember what I said about how direction is important?

You're correct that the other force would be 200N from gravity, but what direction would that force be in?

5. Jun 4, 2014

### ichigo

would the answer to question 1 be 8s?

if t=Δv/acceleration, =80-0/10= 8 seconds?

6. Jun 4, 2014

### ichigo

for question 3: it has a downward force of 200N. The net force would still be 500N right?

7. Jun 4, 2014

Correct.

8. Jun 4, 2014

### ichigo

Okay, my final question:

How do you calculate the distance and displacement on a distance-time graph?

I know that for a velocity-time graph, distance is the area underneath the line, and displacement is the area underneath the line, but taking the negative into account

Is it the same for a distance time graph?

9. Jun 4, 2014

### Nathanael

For a lot of problems you have to choose "the positive direction"

Think of a numberline, the positive direction would be to the right, and so moving to the left would be negative.

In this situation, either up or down would be the positive direction. (Because the problem has nothing to do with left/right, only up/down)

Since the forces are in opposite directions (one up, one down) they will have opposite values (one positive, one negative)

I don't know if I explained this very well but hopefully you understand.

10. Jun 4, 2014

### Nathanael

If you have distance graphed along time (a "distance-time graph") then the displacement will just be the change in the height of the graph.

So say you have the distance as a function of time d(t) (which can be graphed to create a distance-time graph) the displacement from $t_1$ to $t_2$ will be $d(t_2)-d(t_1$)

If you have velocity as a function of time, the displacement from $t_1$ to $t_2$ will be the area under the graph from $t_1$ to $t_2$

But for the distance as a function of time, you just look at the height of the graph, not the area under it.

(I don't know if I explained that clearly, but ask questions if you need to)

11. Jun 4, 2014

### ichigo

does question 3 mean the resultant force? So its 300-200 =100N?

12. Jun 4, 2014

### Nathanael

Yes, the Net force is just the resultant force of ALL the forces acting on something.

13. Jun 4, 2014

### ichigo

I understand everything except for (I'm guessing) the easiest part. How is the distance calculated on a distance time graph? If the line is curving up and down, how would you be able to tell?

14. Jun 4, 2014

### Nathanael

It's an easy concept (the problem is just that it's hard to explain easy concepts)

Basically the distance-time graph is a graph of the displacement over time. So the value (or height) of the graph is the value of the displacement (EXCEPT you also have to subtract the original distance, since "displacement" is "change in distance" which is "final distance minus original distance")

Say at time zero, the distance is also at zero (in other words, the distance-time graph passes through the origin)

Now suppose the graph goes through some crazy wacky curve, but at time 10 seconds, the distance is 100 meters.

The displacement (after 10 seconds) will be 100 meters, regardless of how the graph curves.

Final distance = 100 meters
Original distance = 0 meters (I said that the distance is 0 when the time is 0)

Displacement = 100 - 0 = 100

15. Jun 4, 2014

### ichigo

Oh, sorry, I meant how to calculate the distance OF the line. I understand how to get displacement

16. Jun 4, 2014

### Nathanael

Ok now I don't understand what you mean.

You want the distance of the line? I can't think of any reason you would need to know that (at least not with a distance-time graph)

It depends on the line. A horizontal line (a constant) is trivial.

A straight line (with a slope) is also somewhat trivial (with pythagorean theorem).

Curvy lines are no longer trivial

17. Jun 4, 2014

### ichigo

All good, if that's not useful, then they won't be giving that to me on the exam. Thankyou very much for your help, I really appreciate it!

18. Jun 4, 2014

No problem