Pushing Cart Down a Ramp

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In summary, the professor needs to exert a minimum force of 285 N on the cart to prevent it from exceeding a speed of 7 m/s at the bottom of the ramp. This was calculated using the equations for final velocity, initial velocity, and acceleration, as well as the force equation. The mistake made was corrected and the solution was found to be correct.
  • #1
FuzzyDunlop
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Homework Statement


A professor has to haul a cart down a ramp. The ramp has an angle of about 10 degrees and is about 5 meters long. His initial speed at the top of the ramp is 6 m/s, and the cart has a mass of about 95 kg. How hard does he have to pull on the cart so that at the bottom of the ramp, its speed does not exceed 7 m/s? Neglecting friction, what is the magnitude of the minimum force he has to exert on the cart?

Initial Velocity: 6m/s
Final Velocity: 7m/s
Mass: 95kg
Distance: 5m
Angle: 10

Homework Equations



Vf^2=Vi^2 +2ax
F=ma

The Attempt at a Solution



Well first I found the acceleration of the cart.

49=36 +2a(5)
a=1.3

Then I thought that the net force would be
F=mgsin10 + ma

so

F= 95*9.8*sin(10) + 95*1.3 = 285 N

But I am still getting it wrong. Did I make a mistake someplace?
 
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  • #2
Ohh nevermind I got it. I was right, dumb mistake.
 
  • #3



Your calculations are correct, but you have made a mistake in your application of the equations. The net force on the cart is equal to the sum of the forces acting on it, which includes the force of gravity (mg) and the force needed to accelerate the cart (ma). However, the force needed to accelerate the cart is not equal to the net force, as it also includes the force of friction. Since friction is negligible in this scenario, the force needed to accelerate the cart is equal to the net force, but this is not always the case.

To find the minimum force needed to keep the cart's speed from exceeding 7 m/s, we can use the equation F=ma, where F is the minimum force, m is the mass of the cart, and a is the acceleration. Since we know the final velocity and the distance traveled, we can use the equation Vf^2=Vi^2 +2ax to find the acceleration.

Plugging in the given values, we get:

7^2=6^2 +2a(5)
49=36+10a
a=1.3 m/s^2

Now, we can plug this value of acceleration into the equation F=ma to find the minimum force needed:

F=(95)(1.3)= 123.5 N

Therefore, the professor would need to exert a minimum force of 123.5 N to keep the cart's speed from exceeding 7 m/s.
 

What is the purpose of pushing a cart down a ramp?

The purpose of pushing a cart down a ramp is to demonstrate the principles of force and motion, specifically the effects of gravity and friction on an object's motion.

How does the angle of the ramp affect the cart's motion?

The steeper the angle of the ramp, the faster the cart will accelerate due to the increased force of gravity pulling it downwards. However, if the angle is too steep, the cart may lose control and potentially tip over.

Why does the cart continue to move after it reaches the bottom of the ramp?

The cart continues to move due to inertia, which is the tendency of an object to resist changes in its motion. In this case, the cart is moving in a straight line with a constant velocity until acted upon by another force, such as friction or a barrier.

What factors can affect the distance the cart travels down the ramp?

The distance the cart travels down the ramp can be affected by the angle of the ramp, the mass of the cart, the surface of the ramp (friction), and any external forces acting on the cart (e.g. pushing or pulling it).

How does the weight of the cart affect its motion down the ramp?

The weight of the cart, which is determined by its mass and the force of gravity, will affect the speed and acceleration of the cart down the ramp. A heavier cart will require more force to move and will accelerate slower than a lighter cart.

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