Pulling a crate uphill with friction ()

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In summary: N * 8.04m = 1206J. This energy went into increasing the kinetic energy (KE) and potential energy (PE) of the crate. To calculate the change in KE, we use the formula KE = 1/2mv^2, where m is the mass of the crate (11kg) and v is the final velocity of the crate. We can use the work-energy theorem, which states that the work done is equal to the change in KE, to find the final velocity. So, 1206J = 1/2 * 11kg * v^2, solving for v we get v = 5.19 m/s. This is the final velocity
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killat943
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Homework Statement


A crate is pulled by a force (parallel to the incline) up a rough incline. The crate has an initial speed of 1.39 m/s. The crate is pulled a distance of 8.04 m on the incline by a 150 N force. The acceleration of gravity is 9.8 m/s^2 (The incline is 20 degrees, the coefficient of friction is 0.358 and the weight of the crate is 11 kg.

a) What is the change in the kinetic energy of the crate in units of J?

b) What is the speed of the crate after it is pulled the 8.04 m in units of m/s?

Homework Equations


KE = 1/2mv^2

The Attempt at a Solution



This is the only hint our professor gave us and i feel lost. I just need help setting it up.

we take the system to be the block, then energy is being transferred out of the system (by friction) and into the system (by the applied force F). Setting this up like we set up other conservation of energy problems may be confusing, as you figure out how to deal with the force F. Here's another suggestion: You can calculate how much energy was pulled away from the block by friction and how much energy was added to the block by the force F. When you do that, you will have a number in joules which represents a net gain of energy by the block. That energy had to go somewhere. It went into two places: adding KE to the block and adding PE to the block. You can calculate a number for how much PE was added from the information given. That means that the rest of the energy added went to increasing the KE.

Thanks!
 
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  • #2
Have a go at it - we're eagerly awaiting your attempt so we can make suggestions! At least figure out the friction force, the force of gravity down the ramp - might as well find the acceleration, too. Then you can get the Vf directly as a check on the energy approach.
 
  • #3
Why don't you just try to calculate all those energies like your prof said.

The energy added by a force, also called the work done by the force is force * distance (for a constant force that acts in the direction of movement)
 

1. What is friction and how does it affect pulling a crate uphill?

Friction is a force that opposes the motion of an object. When pulling a crate uphill, friction will act in the opposite direction of the motion, making it more difficult to move the crate.

2. How does the weight of the crate affect the amount of force needed to pull it uphill?

The weight of the crate directly affects the amount of force needed to pull it uphill. The heavier the crate, the more force is needed to overcome the force of friction and move it uphill.

3. Does the surface of the ground impact the amount of friction when pulling a crate uphill?

Yes, the surface of the ground can greatly impact the amount of friction when pulling a crate uphill. Rougher surfaces will create more friction, making it more difficult to move the crate.

4. How can you reduce friction when pulling a crate uphill?

One way to reduce friction when pulling a crate uphill is to use a lubricant, such as oil, to make the surface smoother. Another way is to use wheels or a pulley system to decrease the surface area in contact with the ground, reducing the force of friction.

5. Does the angle of the incline affect the force needed to pull a crate uphill?

Yes, the angle of the incline can greatly impact the force needed to pull a crate uphill. As the angle increases, the force of gravity pulling the crate downhill also increases, making it more difficult to overcome and pull the crate uphill.

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