# Easy physics help

1. Apr 3, 2005

### joejo

Easy physics help!!!

Hi,

1. State the SI units to measure:
a. Work. joules
b. Kinetic Energy. joules
c. Power. watts

2. Calculate the work done by a 140kg black bear that climbs 18m up a tree. 24696J

3. A golf ball is given 115 J of energy by a club that exerts a force over a distance of 4.5cm while the club and the ball are in contact.
a. Calculate the magnitude of the average force exerted by the club on the ball.
b. If the ball’s mass is 47g, find the magnitude of its average acceleration.
c. What speed does the club impart to the ball?

4. A 55kg diver has 1.62 kJ of gravitational potential energy relative to the water when standing on the edge of a diving board. How high is the board above the water? 3.01m

5. Calculate the power of a light bulb that transforms 1.5x104 J of energy per minute. 250w

6. A chair lift takes skiers to the top of a mountain that is 320 m high. The average mass of a skier complete with equipment is 85 kg. The chair lift can deliver three skiers to the top of the mountain every 35s.
a.Determine the power required to carry out this task. (Assume the skiers join the lift at full speed.)
b.If friction increases the power required by 25%, what power must the motors running the lift be able to deliver?

dont understand how to do 6 and 3 and also, if they rest are right?!

Last edited: Apr 3, 2005
2. Apr 3, 2005

### Staff: Mentor

Right.
No. Show your work and don't leave out units.

Hint for a: Consider the work done by the average force.

Right.

Hint for part a: How much work must the lift do against gravity to raise those three skiers?

3. Apr 3, 2005

### HallsofIvy

Staff Emeritus
No, Doc Al didn't say "tell what formulas you used". He said "show your work"- show us your calculations (I get that you are a factor of 10 off).

4. Apr 3, 2005

### Data

Use TeX. You can double click on other peoples' to see the code they used. Here are some examples: https://www.physicsforums.com/showthread.php?t=8997

Anyways, for 2, I get the same result as HallsofIvy: you are an order of magnitude off.

For 4 I want you to look at your answer and tell me if it makes sense. If yes, then post your work. If no, then try to look at your work and fix it, then post it and we'll see if you're right.

For 3a, just apply $\Delta E = F \cdot d$. Try on your own for 3b and 3c~

5. Apr 3, 2005

### whozum

$$W = F * D$$

The SI units for force is $$kg * m / s^2$$ so throughout the entire problem you must use the set of SI units, meters, kilograms, and seconds.

6. Apr 3, 2005

### joejo

for 2....

i used Eg=mgh
=144*9.8*18
=24696J

for 4.....
I used Eg=mgh
1620J=55kg*9.8*h
1620/55*9.8=h
h=3.01m
\for 3.....i got 2555.56N

6. A chair lift takes skiers to the top of a mountain that is 320 m high. The average mass of a skier complete with equipment is 85 kg. The chair lift can deliver three skiers to the top of the mountain every 35s.
a. Determine the power required to carry out this task. (Assume the skiers join the lift at full speed.)
b. If friction increases the power required by 25%, what power must the motors running the lift be able to deliver?

for 6.....
Eg=mgh
=85kg*9.8N/kg*320m
=266560J 2.7*10 to the 5

or

Ek= 1/2mvsquared
= ½(85kg) (9.1m/s)
=3519.43J

for 6.b)

im thinking....
53312J

are any of those answers right.... 3 i dont get....6 i doubt is right!!!

Last edited: Apr 3, 2005
7. Apr 4, 2005

### whozum

For 6, You travel 320m against gravity, whos force is equal to 85*3*9.8. That is the work done. Power is work done divided by time interval. Your answer should be about 7500W.

If friction requires 25% more power than provided, then find 25% of the power and add it to the given power.

Note that 9.1m/s is not the velocity of the ski lift is not 320/35. You don't know the total distance traveled as it is irrelevant, the only thing that matters is the work done parallel to the force of gravity, in which case 320m worth of work.

8. Apr 4, 2005

### joejo

hey,
so are #'s 1-5 right and just 6 wrong! As i redid some of them....thanks again

9. Apr 4, 2005

### Staff: Mentor

OK.

OK.
That's good for 3a.
That's the work needed to raise one skier; now calculate the power needed if three skiers are raised in the given time.

This method doesn't apply for several reasons. For one thing, as whozum points out, 9.1 m/s not the correct speed of the lift (it's just the vertical component). More important, the speed of the lift is assumed to be constant. Note where is says "Assume the skiers join the lift at full speed". So the lift does not change the kinetic energy of the skiers, only the potential energy.

10. Apr 4, 2005

### joejo

alright i've been trying for the past few hours, and untill now I don't get 3.b, c and 6.......can someone please explain...i've spent more then enough time and nothing!! I'm homeschooling so I teach myself...thats why im always on forums asking for help!! thanks guys

11. Apr 4, 2005

### joejo

anyone!!!!

12. Apr 4, 2005

### whozum

I think we've explained 6 pretty well, tell us what you dont get.

As for 3 you are given:

W = 115J
d = 4.5cm = 0.045m
Find F

Well the only equation relating these 3 is W = F * d. From there, a) is trivial.

b) Find mass of the ball using the force from a) and mass 47g

F = ma, a = F/m
m = 47g = 0.047kg
F = from a)

c) The energy imparted on the ball was 115J, this is the amount of energy the ball has, and since it is all kinetic, use the equation for kinetic energy to find the speed, since you know the mass.

13. Apr 5, 2005

### joejo

i still dont get 6....3 i understand now!!! someone please show me how to do it! thanks

14. Apr 5, 2005

### whozum

6. A chair lift takes skiers to the top of a mountain that is 320 m high. The average mass of a skier complete with equipment is 85 kg. The chair lift can deliver three skiers to the top of the mountain every 35s.
a. Determine the power required to carry out this task. (Assume the skiers join the lift at full speed.)
b. If friction increases the power required by 25%, what power must the motors running the lift be able to deliver?

Imagine a ski lift. Draw a triangle, the vertical end is 320m, the hypotenuse is the path of the lift. Draw a force vector pointing down on the vertical end, label this gravity with magnitude m*9.8. The direction of travel is along the hypotenuse, but recall that the work done against gravity is the dot product of the gravitational force and the path of travel.

$$W = \int F \bullet dx = \int Fcos(\theta) dx = Fxcos(\theta)$$

The dot product will find you the component of travel parallel to the force, which is what matters. In the problem, the componenet parallel to the force is already given, it is 320m. If the height of the triangle is 320m, you know he travelled a distance much longer, since the hypotenuse is always larger than either of its legs in a right triangle. Given the displacement AGAINST gravity, and the gravitational force:

$$W = F d$$. There is no cos theta since you have already found the parallel component, all the hard work is done. Now just plug and chug:

$$W = F d, F=85*3*9.8 = 2499, and d = 320m$$

$$W = (2499)(320) = 799680J$$

Power is the Work done in the time interval,

$$P = \frac{W}{t}$$

Well we just figured out the work, and we know it was done in 35seconds so:

$$P = \frac{W}{t} = {799680}{35} = 22848W$$

15. Apr 5, 2005

### joejo

ohh okay...now i get it......thanks alot whozum.....

now for 3....

i got a) 2555.6N using W=F*D b) I used a=F/m and I got 54374.5 c)???? I dont get....do I use E=1/2mvsquared...using the 47g...because it asks what speed does the club imart to the ball...and so wouldn't i need the mass of the club??

16. Apr 5, 2005

### whozum

No, notice they already gave you the energy imparted on the ball. Whatever mass or velocity the club was swung at is irrelevant, since you would use that information to find the energy imparted. Once you have that, just apply that energy to the ball.

Your figures are right. Note that thats alot of energy for such a small ball, so it makes sense that it would go pretty fast.

17. Apr 5, 2005

### joejo

for c) im thinking 69.95m/s

18. Apr 5, 2005

### joejo

so would 69.96m/s be right....and for the units on 3.B) would it be 54374.5m/s²

19. Apr 5, 2005

### whozum

$$W = 115J, m = 47g = 0.047kg$$

$$W = 1/2 mv^2$$

$$v = \sqrt{\frac{2W}{m}} = \sqrt{\frac{230}{0.047}} = 69.95m/s$$

The units of acceleration are m/s^2

20. Apr 5, 2005

### joejo

thanks alot whozum!! your the best!!