# Acceleration on a slant/coefficient of friction

• Crazdfanatic
In summary, the problem involves a block with an initial speed of 6.00 m/s sliding up an inclined plane at an angle of 30 degrees with the horizontal. The coefficient of kinetic friction is 0.330 and the goal is to find the maximum distance the block will travel up the incline. After attempting to use the equation delta_x = v_initial + .5a*t^2 and adjusting for the slanted surface, the student was unable to find the correct answer. Upon further explanation of the coefficient of friction and the use of normal force in relation to the angle of the incline, the student attempted the problem again with a new acceleration of -2.80 m/s/s. However, their calculations still did not match
Crazdfanatic

## Homework Statement

A block is given an initial speed of 6.00 m/s as it slides up an inclined plane that makes an angle of 30 degrees with the horizontal. If the coefficient of kinetic friction is .330, how far up the incline will the block go?

## Homework Equations

I messed around with the equation delta_x = v_initial + .5a*t^2 , not sure how the coefficient of friction works

## The Attempt at a Solution

I tried setting a = - 11.3 m/s/s by using trig because of the slant on the horizon, then using the above distance formula to find the maximum distance it traveled. I tried multiplying the coefficient of friction into the acceleration part of the equation, but to no avail. Please help me figure out what I'm doing wrong. Maybe an explanation of coefficient of friction is all I really need, I don't know.

The coefficient of friction defines the relation between the normal force (exerted perpendicular to the contact surface) and the frictional force (which is parallel to the surface).

Ffrict = u x Fnormal

On a ramp, the normal force is no longer equal to simply mg, which is where the angles come in. Give it another try using these principles.

Okay, so I looked at it again, and this is what I got.

I found the new acceleration by g(u)cos(30) . I ended up getting a = -2.80 m/s/s.

From there, I plugged it into the distance formula above, leaving me with:

delta_x = t(6.00) + .5(-2.80)t^2

I took the derivative with respect to t to find the t that would give me the max distance.
I got this.

0 = 6.00 + (-2.80)t
t = 2.14 s

I plug that back into the above equation to get delta_x.

I ended up getting delta_x = 6.42 m, which is nowhere near any of the answer choices.

Am I even on the right track?

## What is acceleration on a slant?

Acceleration on a slant refers to the rate of change of an object's velocity as it moves along a slanted surface. It is affected by the angle of the surface as well as the force acting on the object.

## How is acceleration on a slant calculated?

Acceleration on a slant is calculated using the formula a = gsinθ - μgcosθ, where a is the acceleration, g is the gravitational constant, θ is the angle of the surface, and μ is the coefficient of friction.

## What is the coefficient of friction?

The coefficient of friction is a dimensionless constant that measures the amount of friction between two surfaces in contact. It is determined by the type of materials and the force pushing them together.

## How does the coefficient of friction affect acceleration on a slant?

The coefficient of friction plays a significant role in determining the acceleration on a slant. A higher coefficient of friction means there is more resistance to motion, resulting in a lower acceleration. Conversely, a lower coefficient of friction leads to a higher acceleration.

## What is the relationship between acceleration on a slant and the angle of the surface?

The angle of the surface has a direct impact on the acceleration on a slant. As the angle increases, the acceleration decreases, reaching zero at a 90-degree angle. This is because a steeper angle increases the force of gravity acting on the object, while also increasing the resistance due to the normal force.

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