How High Does the Crate Reach on the Inclined Plane?

[Leesh09]

Homework Statement

1. A crate with scientific equipment slides down a curved frictionless side of lunar crater of the depth h, and then up along the other side which is an inclined plane as in Figure 2. The coefficient of kinetic friction between crate and incline is k, and the inclined side makes an angle θ with the horizontal. Use energy methods to find the maximum height ymax reached by the crate.

Homework Equations

The Attempt at a Solution

The equation I have come up with so far is mgy=mgh-(Mu sub k *n)y/sin(theta)

I know this is correct, but how would one calculate what the normal force would be? or is it sufficient to simply solve for y?

 

[tiny-tim]

Hi Leesh09!

(btw, you shouldn't really call it g should you? )

… how would one calculate what the normal force would be? or is it sufficient to simply solve for y?

Yes, you do need to know N.

Hint: there's zero acceleration normal to the incline.

 

[tomcue]

I would first clarify the problem by checking the given values for the mass of the crate (m), the coefficient of kinetic friction (k), and the angle of the inclined plane (θ). I would also visualize the problem to better understand the situation.

Next, I would approach the problem using the principle of conservation of energy, which states that the total energy of a system remains constant. In this case, the initial potential energy of the crate at the top of the lunar crater is converted into kinetic energy as it slides down, then back into potential energy as it reaches the maximum height on the other side of the inclined plane. This can be expressed mathematically as:

mgh = 1/2mv^2 + mghmax

Where m is the mass of the crate, g is the acceleration due to gravity, h is the initial height of the crate, v is the velocity of the crate at the bottom of the crater, and hmax is the maximum height reached by the crate on the inclined plane.

To find hmax, we can use the equation for kinetic friction to find the work done by friction on the crate as it slides up the inclined plane:

Wfric = -Mu sub k *n *d

Where Wfric is the work done by friction, Mu sub k is the coefficient of kinetic friction, n is the normal force, and d is the distance traveled. We can then equate this work to the change in potential energy of the crate:

Wfric = mghmax - mgh

Substituting in the value for the normal force as n = mgcosθ, we can solve for hmax:

hmax = h + d - (Mu sub k *d)/cosθ

Finally, to find the maximum height reached by the crate (ymax), we can use the fact that the distance traveled along the inclined plane (d) is equal to hmax/sinθ, giving us the final equation:

ymax = h + (hmax - h)/sinθ - (Mu sub k *hmax)/sinθ

In conclusion, using energy methods, we can find the maximum height reached by the crate as it slides down and up the lunar crater. It is important to note that this solution assumes a perfectly frictionless curved side of the crater, which may not be realistic in a real-world scenario.