I was wondering if my answers are correct.
1. A 50kg person is taking a ride on an elevator travelling up at a steady speed of 2.5m/s. Find the time fo the elevator trip if the elevator does 4900 J of work on the person.
My answer: First I used the formula f=mg to find the force and got -490N. Then, I used the firmula w=fd and found the distance which is 10m. Lastly, I used v=d/t to find the time and got -4.0 seconds. Time shouldnt be negative but I got a negative number from the force equation.
2. A car slows from 108km/h to 90km/h over a course of 4.0s. THe force of the brakes was 1000N. What is the work done by the brakes on the car?
My answer: 110 000N
3. A pair of sleds weighs 25 kg and experience frictional force of 25N. If a dog team applies a 175 N force pulling it 12m find:
a) work done by friction
b) work done by dog team
c) net work done (using net force). how does the net work compare to answers from a and b
d) what form of energy does the frictional force represent? what form of energy does the net force produce?
My answers:
a) 300 N•m
b) 2100 N•m
c) net force=200N and net work=2400N•m. I'm not sure how it would compare to the previous answers.
d) frictional force represents thermal energy and net force produces kinetic energy?
for question number 1
we need to find time but we cant find time if we don't have the distance. So first we will find distance.So apply a=Fnet/m ,
a=0(speed is constant).
0*50=Fnet=0N
then as we have the mass the other force applying here is the gravitational force
so Fnet=Fapp-Fg(gravitational force is acting in the opposite direction)
We will find Fg=mg,
Fg=50*9.8=490N
so Fnet=Fapp-Fg
0N=Fapp-490
4900N=Fapp
W=Fapp*d(work is the product of force applied and distance covered)
4900=490d
4900/490=d
10m=d
No we have distance and velocity we can easily find time
v=d/t
t=d/v
t=10/2.5
t=4seconds
2. A car slows from 108km/h to 90km/h over a course of 4.0s. THe force of the brakes was 1000N. What is the work done by the brakes on the car?
My answer: 110 000N
+++++++++++++++++++++++++++++++++++++++
How far?
In constant acceleration, average velocity is (final +initial)/2
v = 99 *1000/3600 = 27.5 m/s
goes 4 s * 27.5 m/s = 110 meters
1000N * 110 m = 110,000 Joules
So YES
a) 300 N•m YES (or JOULES)
b) 2100 N•m YES (or Joules
c) net force=200N and net work=2400N•m.
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I get + for pull and - for friction so
net = 175 - 25 = 150N
Net work = 150 * 12 = 1800 Joules
and
1800 = 2100 -300 indeed
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I'm not sure how it would compare to the previous answers.
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useful work = work done - heat generated
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d) frictional force represents thermal energy and net force produces kinetic energy? yes
Ohhh I see what I did wrong! Thanks a lot again!
1. Wow, it seems like you've done some heavy lifting (or rather, heavy calculating) with this question! While your approach is commendable, there seems to be a slight mishap in your calculations. When calculating the force, make sure to take the magnitude and direction into account. Since the elevator does positive work on the person, the force should be positive as well. That means you'll end up with a positive distance and positive time, which is much more cheerful!
2. Oh brakes, always coming to a stop and spoiling the fun! In this case, to calculate the work done by the brakes, you'll want to use the work formula, which is W = Fd (not N, by the way). Remember to convert the speeds from km/h to m/s first. Once you crack that secret code, you should be able to calculate the work done by those brake-happy brakes!
3a. Ah, friction, always trying to slow things down! To calculate the work done by friction, you'll again use the trusty work formula. Remember that the work done by a force is equal to the force multiplied by the distance it acts over (W = Fd). So with that in mind, let the sleds slide into the exciting world of frictional work!
3b. Now it's time for the dog team to shine! They're pulling those sleds like champs, and we want to know their work. Again, use the work formula (W = Fd), but this time plug in the force applied by the dog team and the distance they pull those sleds. Let the doggos show off their work!
3c. Ah, net force, the superhero that saves the day! To find the net work done, you'll want to take into account both the work done by friction and the work done by the dog team. Add those two values together, and voila! You'll have the net work done. Compare it to the individual works and see who the real hero is in this work-filled saga!
3d. Ah, the energy question, always a classic! The frictional force represents thermal energy, as it generates heat through the friction between the sleds and the ground. On the other hand, the net force produces the kinetic energy of the sleds as they zoom through the snowy landscape. Quite a polar opposite, wouldn't you agree?
Let's go through each question to see if your answers are correct.
1. For the first question, you correctly used the formula F = mg to find the force acting on the person as -490N (negative because it's in the opposite direction to the motion). However, it seems there was a mistake in your calculation of the distance. The work done by the elevator can be calculated using the formula W = Fd, where W is the work done, F is the force, and d is the distance. Rearranging this formula, we have d = W / F. Plugging in the values, we get d = 4900 J / -490 N = -10 m. Since distance cannot be negative, you should take the magnitude of the distance, which is 10 m. Now, you can use the formula v = d / t to solve for time. Rearranging the formula, we have t = d / v = 10 m / 2.5 m/s = 4 seconds. So, the correct answer is 4 seconds.
2. For the second question, you've mentioned the force of the brakes as 1000N. However, to find the work done by the brakes, you need to use the formula W = Fd, where W is the work done, F is the force applied, and d is the distance over which the force is applied. The work done by the brakes would be W = 1000 N * d. We don't have the distance mentioned in the problem, so it's not possible to calculate the work done by the brakes.
3. For the third question, let's analyze each part:
a) The work done by friction can be calculated using the formula W = Fd, where W is the work done, F is the frictional force, and d is the distance. Plugging in the values, we get W = 25 N * 12 m = 300 N·m.
b) The work done by the dog team can be calculated using the same formula. Plugging in the values, we get W = 175 N * 12 m = 2100 N·m.
c) The net work done is the sum of the work done by both the frictional force and the dog team. So, net work = 300 N·m + 2100 N·m = 2400 N·m. The net force can be calculated using Newton's second law: net force = mass * acceleration. Since the sleds experience a net force of 175 N and they have a total mass of 25 kg, the acceleration can be found using the formula acceleration = net force / mass = 175 N / 25 kg = 7 m/s^2. Now, you can use the formula W = (1/2) * m * v^2 to find the net work done. Here, v is the final velocity, which can be calculated using the formula v = u + a * t, where u is the initial velocity, a is the acceleration, and t is the time taken. Since u is not given, we assume it to be 0 m/s. Plugging in the values, we get v = 0 m/s + 7 m/s^2 * 4 s = 28 m/s. Now we can calculate the net work done as W = (1/2) * 25 kg * (28 m/s)^2 = 9800 N·m.
Comparing the answers from parts a) and b) to the net work done in part c), we can see that the net work done is equal to the sum of the work done by the frictional force and the dog team.
d) The frictional force represents the conversion of mechanical energy (kinetic energy) into thermal energy due to friction. The net force, being responsible for the acceleration of the sleds, produces kinetic energy.
Based on this explanation, it seems that your answers for parts a), b), and d) are correct, but the answer for part c) is incorrect and needs to be recalculated.
1. A 50kg person is taking a ride on an elevator travelling up at a steady speed of 2.5m/s. Find the time fo the elevator trip if the elevator does 4900 J of work on the person.
My answer: First I used the formula f=mg to find the force and got -490N.
************
- 490 N is gravity force DOWN on person.
+ 490 N is force elevator exerts UP on person
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Then, I used the firmula w=fd and found the distance which is 10m.
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YES
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Lastly, I used v=d/t to find the time and got -4.0 seconds. Time shouldnt be negative but I got a negative number from the force equation.
+++++++++++++++++++++
Well my force is POSITIVE :)
+++++++++++++++++++++++ So I get +4 sec