How Much Work Does It Take to Pull a Skier Up a Slope?

[scurry18]

Homework Statement

A skier of mass 70.0 kg is pulled up a slope by a motor-driven cable.
How much work is required to pull the skiera distance of 60.0m up a 30 degree slope (assumed frictionless) at a constant speed of 2.00 m/s?

Homework Equations

W=F (delta)r cos(theta)

The Attempt at a Solution

(delta)r= 60.0m
(theta)=30.0 degrees
Force=60sin30=30 J


Is this the correct start?

 

[Hootenanny]

Homework Statement

A skier of mass 70.0 kg is pulled up a slope by a motor-driven cable.
How much work is required to pull the skiera distance of 60.0m up a 30 degree slope (assumed frictionless) at a constant speed of 2.00 m/s?

Homework Equations

W=F (delta)r cos(theta)

The Attempt at a Solution

(delta)r= 60.0m
(theta)=30.0 degrees

It's all good up until this point:

Force=60sin30=30 J

Firstly force is measured in Newton's and not Joule's. Secondly, how did you calculate the force acting on the skier?

HINT: This problem is much more straight forward if one simply considered the gravitational potential energy of the skier.

 

[baba89]

Yes, this is a good start! You correctly identified the necessary information, such as the mass of the skier (70.0 kg), the distance traveled (60.0 m), and the angle of the slope (30 degrees). You also correctly identified the formula for work, which includes the force (F), distance (delta)r, and the cosine of the angle (cos(theta)).

However, there are a few things to note. First, the formula for work is actually W=F(delta)r, not W=F(delta)r cos(theta). This is because the cosine of the angle only comes into play when the force is not parallel to the direction of motion. In this case, the force pulling the skier (provided by the motor-driven cable) is parallel to the direction of motion, so we do not need to include the cosine term.
Second, the force you calculated (30 J) is actually the magnitude of the force, not the actual force. Remember, force is a vector quantity and has both magnitude and direction. In this problem, the force is acting in the same direction as the motion of the skier, so the actual force would be positive 30 J.
Finally, to calculate the work, you would need to multiply the force by the distance traveled, so the final answer would be W=(30 J)(60.0 m)=1800 J.

I hope this helps clarify the solution for you! Keep up the good work.