Kinetic and Potential Energy on Ramps

[captdestiny]

Homework Statement

I don't want to tell the problem I just want to know how to solve it.
There is a mass and a velocity and the ramp is inclined let's say 30 degrees.
The object is going up the ramp with a force of friction opposing it. Find the distance the object travels before stopping.

Homework Equations

W = F (d)
Eg = mgh
Ek = 1/2(mv^2)

The Attempt at a Solution

let m = 5
let v = 10


Ek = 1/2(5(10^2))
= 250 J

W = 250 J
250J = 4(d) ?
62.5 = d ?


Something wrong I am doing, all help is gratefully apppreciated.

 

[BetoG93]

Part of the kinetic energy is lost due to friction, and part is converted to gravitational potential energy.
Ef - Ei = -Fd = mgh - (1/2)mv^2

 

[netgypsy]

Yes you did forget to subtract the frictional work from the original kinetic energy but once you find the vertical height the object rises you have to use the given angle to find how far up the ramp it will go.

 

[netgypsy]

Yes you do subtract and then set the answer equal to mgh, the potential energy and then solve for the vertical height. But this isn't the answer to the question. The object moves up a ramp, not straight up so you have to draw the triangle showing the vertical height which you now know, the angle the ramp makes with the horizontal, which you given and then use the appropriate trig function to solve for the ramp which is the hypotenuse of the triangle.

 

[asteriatic]

To solve this problem, we can use the conservation of energy principle, which states that the total energy of a system remains constant. In this case, the total energy is the sum of the object's kinetic energy (Ek) and potential energy (Eg).

First, we can calculate the potential energy of the object at the bottom of the ramp using Eg = mgh, where m is the mass, g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the ramp. Since the object is at the bottom of the ramp, h = 0, so Eg = 0 J.

Next, we can calculate the kinetic energy of the object using Ek = 1/2(mv^2), where m is the mass and v is the velocity. Since the object is moving up the ramp, we need to use the component of the velocity that is parallel to the ramp. This can be found using the formula v = v0sinθ, where v0 is the initial velocity and θ is the angle of the ramp (30 degrees in this case). So, v = 10sin(30) = 5 m/s. Plugging this into the kinetic energy equation, we get Ek = 1/2(5(5^2)) = 62.5 J.

Now, we can use the conservation of energy principle to find the distance traveled by the object before it stops. Since the total energy remains constant, we can equate the initial potential energy (Eg) to the final kinetic energy (Ek). So, mgh = 1/2(mv^2). Rearranging this equation, we get h = 1/2(v^2/g). Plugging in the values, we get h = 1/2(5^2/9.8) = 1.275 m.

This means that the object will travel a distance of 1.275 meters up the ramp before it stops. However, we also need to take into account the force of friction opposing the object's motion. This force will decrease the object's kinetic energy and eventually bring it to a stop. The exact distance traveled will depend on the magnitude of the force of friction, which is not given in the problem.

In conclusion, to find the distance traveled by the object on the ramp, we need to use the conservation of energy principle and take into