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Calculate the work done by friction as a 3.2 kg box is slid along a floor from point A to point B in the figure along paths 1, 2, and 3. Assume that the coefficient of kinetic friction between the box and the floor is 0.26.

Path 1: Extends 4 m up from point A, then 4 m to the right, then 1 m down, then 1 m left, then 1 m down.
Path 2: Extends 2 m right from point A, then up 2 m, then right 1 m.
Path 3: Extends 1 m down from point A, then right 3 m, then up 3 m.

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Answer:

The work done by friction along paths 1, 2, and 3 are 89.76J, 40.8J, and 57.12J respectively.

Explanation:

To calculate the work done by friction along each path, you first need to find the total distance traveled by the box and then use the work-energy principle, which states that work done is equal to the change in kinetic energy.

Let's calculate the work done along each path:

Path 1: Total distance = 4m + 4m + 1m + 1m + 1m = 11m

Work done by friction = force of friction x distance Force of friction = coefficient of kinetic friction x normal force Normal force = weight of the box = mass x gravity Normal force = 3.2kg x 9.8m/s^2 = 31.36N

Force of friction = 0.26 x 31.36 = 8.16N

Work done by friction = force of friction x distance = 8.16N x 11m = 89.76J

Path 2: Total distance = 2m + 2m + 1m = 5m

Work done by friction = force of friction x distance Force of friction = 0.26 x 31.36 = 8.16N

Work done by friction = 8.16N x 5m = 40.8J

Path 3: Total distance = 1m + 3m + 3m = 7m

Work done by friction = force of friction x distance Force of friction = 0.26 x 31.36 = 8.16N

Work done by friction = 8.16N x 7m = 57.12J

Therefore, the work done by friction along paths 1, 2, and 3 are 89.76J, 40.8J, and 57.12J respectively.

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