EK 1001 Physics 448 Ramp/Mechanical Advantage

[GRod18]

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I just can't find the answer. EK said that there are way too many calc on this problem than for the MCAT, but its just fraustrating that I still can't find the right answer with their method. 😡

448. a farmer has a 200 kg barrel that he must pull up a ramp to a height of 5 m. His old horse can apply a maximum force of only 1,000 N and only for 4 S. what is the shortest distance that the farmer can make the ramp if the barrel starts from rest.

Answer 20 meters.

Thanks in advance!

 

[Rabolisk]

What did EK use? I used kinematics..

Draw a ramp, where the hypotenuse is the unknown x, and the height is 5. Since the horse can only apply force for 4 seconds, that means that in 4 seconds, the entire distance x must be traveled, setting up the equation

x = 1/2at^2, t = 4
Solving for a gives x/8. The sum of all forces acting on the barrel is equal to its mass times acceleration, F = ma. The horse provides a constant force of 1000N, and gravity provides some force down the ramp. This gravitational force is mgsinθ. Looking at our diagram, sinθ is 5/x. Our equation is now ma = 1000 - 5mg/x. Substitute in for a, m, and g yields 25x = 1000 - 10000/x. Multiply all terms by x, and use the quadratic formula and you get x = 20.

 

[GRod18]

What did EK use? I used kinematics..

Draw a ramp, where the hypotenuse is the unknown x, and the height is 5. Since the horse can only apply force for 4 seconds, that means that in 4 seconds, the entire distance x must be traveled, setting up the equation

x = 1/2at^2, t = 4
Solving for a gives x/8. The sum of all forces acting on the barrel is equal to its mass times acceleration, F = ma. The horse provides a constant force of 1000N, and gravity provides some force down the ramp. This gravitational force is mgsinθ. Looking at our diagram, sinθ is 5/x. Our equation is now ma = 1000 - 5mg/x. Substitute in for a, m, and g yields 25x = 1000 - 10000/x. Multiply all terms by x, and use the quadratic formula and you get x = 20.

Your a life saver Rabolisk,

Ek solved for net force up the ramp Fnet = F-mgsin(theta) = ma d = 1/2 at^2 their final equation was d = 1/2 (F-mgh/d / m) t^2. I plugged in values couldn't get the answer.