- #1

jfnn

## Homework Statement

A box slides on a smooth (frictionless) horizontal surface, 1.50 m above the floor, at a speed of 2.00 m/s. The box then slides down a ramp that makes an angle of 36 deg with the horizontal and has a coefficient of kinetic friction equal to 0.430. After reaching the end of the ramp, the box continues to slide horizontally on a smooth ground. Use the law of conservation of energy to calculate the final speed (v) of the box as it slides on the floor.

## Homework Equations

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KEi + PEi = KEf + PEf

and

KEi + PEi + Wnc = KEf + PEf

## The Attempt at a Solution

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I broke it into three segments. I first did the flat floor above the ramp. Since it is all conservative forces acting, I used KEi + PEi = KEf + PEf. Furthermore, down the ramp i defined as negative and south is negative.

Therefore,

1/2m*vi^2 + mghi = 1/2m*v^2 = mgh

In the above equation, the height initial (hi) and the final height (h) are both the same. So the mgh will cancel leaving

1/2m*vi^2 = 1/2m*v^2

After simplifying,

vi^2=v^2 --> Since vi is -2.00 m/s (it is going down in the direction towards the ramp, which is to the left_

(2.00)^2 = v^2

4=v^2

v=-2.00 m/s --> This will now become the initial velocity for the segment involving the ramp.

Since the ramp has friction, we need to consider work by the non-conservative forces -->

KEi + PEi + Wnc = KEf + PEf

1/2m*vi^2 + mghi + Wnc = 1/2m*v^2 + mgh

Where: The initial velocity (vi) = -2.00 m/s, the initial height (hi) is 1.5m, the final height (h) is 0, and wnc= dfcos(theta).

1/2m*(-2)^2 + m(9.8)(1.5) + d*friction*cos(theta) = 1/2m*v^2 + m(9.8)(0)

Therefore, 16.7m - d*uk*N = 1/2m*v^2

The distance (d) is: sin(36) = 1.5 --> d = 2.5519m

16.7m - (2.5519m)(0.430)*N = 1/2m*v^2

16.7m - 1.0973N = 1/2m*v^2

I use a force diagram of the box on the ramp, and I see that N=w perpendicular --> N= mgcos(36)

33.4m - mgcos(36) * 2.1946 = v^2

33.4 - 17.3995 = v^2

16=v^2

v= +/- 4.00 m/s --> I use negative value, again because I defined down the ramp as negative direction for x

Finally, just to show that this is the final velocity of the box, I take the segment of the box just hitting the ground where only conservative forces act, to show that the final velocity as it is sliding on the floor is -4.00 m/. This question asks one for speed, which is scalar, so the answer is 4.00 m/s. Is this the correct logic for a problem like this? Could someone verify the answer?

Thank you in advance.