- #1

learnitall

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## Homework Statement

A box is sliding down an incline tilted at a 11.1° angle above the horizontal. The box is initially sliding down the incline at a speed of 1.70 m/s. The coefficient of kinetic friction between the box and the incline is 0.390. How far does the box slide down the incline before coming to rest.

a) 1.08 m

b)0.775 m

c)0.620 m

d)0.929 m

e)The box does not stop. It accelerates down the plane

## Homework Equations

**Work-Energy Therorem:**

**U1 + K1 + W**

_{other}= U2 + K2**K=(1/2)mv**

^{2}**U=mgh**

**Work=Fdr**

**F**

_{friction}=μF_{normal}## The Attempt at a Solution

Given an initial velocity and a coefficient of kinetic friction, immediately I thought of using the work energy theorem to solve this problem.

I first set up a coordinate system that is tilted and running along the plane, with the origin at the top of plane.

I assumed the block starts from the top (with given initial velocity) and ends at a distance d - the unknown variable we're solving for.

I drew a free body diagram of the box and summed the forces in both the X an Y directions.

Here is the result of that:

**ƩF**

ƩF

_{x}= mgsinθ - F_{friction}ƩF

_{y}=F_{normal}-mgcosθ=0I ignored the Fy equation and used the Fx one to get:

**F**

_{normal}=mgcosθso now I set up my Work-Energy Theorem Equation exactly like how I wrote it in part 2 above:

**0 + (1/2)mv**

^{2}- μmgcosθd = 0 + 0The equation is set to zero because the block stops at the end of the interval (∴ K2= 0) and there is never potential energy as per my coordinate system (∴U2=0)

I solved my Work-Energy equation for 'd' and got 0.3852819047≈0.385.

Can anyone tell me where I went wrong? Did I take the wrong approach? which one should I take?

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