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DedPeter [7]
3 years ago
13

You pull a 70-kg crate at an angle of 30° above the horizontal. If you pull with a force of 600N and the coefficient of kinetic

friction is .4, how far will it move in 3 seconds?
Physics
1 answer:
Alenkinab [10]3 years ago
5 0

Answer:

Explanation:

Force of friction acting on the body = μ mg cosθ

= .4  x 70 x 9.8 x cos30

= 237.63 N

component of weight = mgsinθ

= 70 x 9.8 x sin30

= 343 N  

Net upward force = 600 - mgsinθ - μ mg cosθ

= 600 - 343 - 237.63

= 105.37 N

acceleration in upward direction = 105.37 / 70

= 1.5 m /s²

s = ut + 1/2 a t²

= 0 + .5 x 1.5 x 3²

= 6.75 m .

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In the experiment the students indicate that they can measure the position, velocity and acceleration of the body.

Based on the above, if we place several filter weights and measure their speed and acceleration in each case, we can make a graph of velocity versus acceleration, we can take the value of the constant b from the slope.

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B) False. The constant depends on the area, so it must be kept constant.

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D) True. According to the discussion of the velocity versus acceleration. graph, the constant b is equal to the slope of the graph.

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In conclusion, using Newton's second law and graphical analysis we can find the correct answer for the measure of the constant b is:

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A 0.5 kg mass on a spring undergoes simple harmonic motion with a total mechanical energy of 12 J. If the oscillation amplitude
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Answer:

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