Work & Horsepower: Pulling a 65 kg Skier Up a 30° Slope

  • Thread starter Thread starter netprincess
  • Start date Start date
  • Tags Tags
    Horsepower Work
AI Thread Summary
To calculate the work required to pull a 65 kg skier up a 30° slope for 50 m at a constant speed of 2.9 m/s, one must first determine the force needed to counteract gravity on the slope. The vertical component of the skier's weight can be calculated to find the necessary pulling force. The work done is then found using the formula W = Fd, where F is the force and d is the distance. Additionally, to find the horsepower required, the work done over time must be converted to power, noting that 1 horsepower equals 746 watts. The discussion emphasizes understanding the relationship between force, work, and power in this frictionless scenario.
netprincess
Messages
1
Reaction score
0
A skier of mass 65 kg is pulled up a slope by a motor-driven cable, a)How much work is required to pull him 50 m up a 30° slope (assumed frictionless) at a constant speed of 2.9 m/s? b)how much horsepower must the motor have to preform this task?

Please help! I don't know what to do with this... I know it's a simple question but I've managed to get all confused. Where do I start? :rolleyes: Thank you!
 
Physics news on Phys.org
Find the force pulling the skier down the hill. This is the force required to pull him up. That is a good place to start. From there, remember that KE=(1/2)mv^2 and also that W=Fd.
 
netprincess said:
A skier of mass 65 kg is pulled up a slope by a motor-driven cable, a)How much work is required to pull him 50 m up a 30° slope (assumed frictionless) at a constant speed of 2.9 m/s? b)how much horsepower must the motor have to preform this task?
a) How long does it take to cover 50 m if the speed is 2.9 m/sec? How much vertical distance does it rise in that time? How much energy did it use in that time? What is the rate of work done/unit time?

b) How many Joules/sec is 1 horsepower?

AM
 
Mr. Snookums said:
Find the force pulling the skier down the hill. This is the force required to pull him up. That is a good place to start. From there, remember that KE=(1/2)mv^2 and also that W=Fd.
There is no change in kinetic energy in the problem as stated, so the power is not used to provide kinetic energy.

AM
 
Andrew Mason said:
There is no change in kinetic energy in the problem as stated, so the power is not used to provide kinetic energy.

AM

If it has velocity and mass, doesn't it have kinetic energy?
 
The skier does have kinetic energy when he's being pulled. I think what Andrew meant was that there was no change in this KE, as the question didn't mention about the skier starting from rest, but being pulled with constant velocity instead.
 

Thread 'Errors and significant figures'

I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
View full post »
Thread 'Collision of a bullet on a rod-string system: query'

Thread 'Collision of a bullet on a rod-string system: query'

In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
View full post »
Thread 'A cylinder connected to a hanging mass'

Thread 'A cylinder connected to a hanging mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
View full post »

Similar threads

Skiing Up a Slope: Solving Work & Power Requirements
Replies
9
Views
2K
Work and Energy on a Slope - How Does a Block Move Up with Zero Net Work?
Replies
3
Views
2K
Power Needed to Pull Skier Up Slope
Replies
1
Views
2K
How Much Work Does It Take to Pull a Skier Up a Slope?
Replies
1
Views
2K
Solve a Skier Work Physics Problem on a 37.2° Slope – No Friction
Replies
3
Views
2K
What is the average force required to pull the rope?
Replies
36
Views
6K
The Energy and Efficiency of Cross-Country Skiing: A Mathematical Analysis
Replies
3
Views
3K
How Much Work is Needed to Pull a Skier Up a Slope?
Replies
1
Views
2K
Calculating Work to Pull a Skier Up a Slope at Constant Speed
Replies
2
Views
2K
What's the source of increase in rotational energy of carousel?
Replies
9
Views
2K

Hot Threads

Recent Insights

Back
Top