# Physics!

HELP!!!

A box rest on an incline making a 34 angle with the horizontal. It is found that a parallel force to the incline of at least 240 N can prevent the box from sliding down the incline. If the weight of the box is 800 N, find the coefficient of static friction between the box and the incline.

1. weight component down = m g cos 34
friction force up = m g cos 34 * mu
so
mu m g cos 34 + 240 = m g sin 34

posted by Damon
2. I get 8.01. That doesn't seem correct. I plugged in all my numbers. Anything you think I could have done incorrectly?

posted by Randy
3. I converted newtons to kilograms and my answer makes more sense to me. I got 0.312 for mu this time. Thank you for the guidance.

posted by Randy
4. M*g = 800 N.

Fp = 800*sin34 = 447.4 N.

Fn = 800*Cos34 = 663.2 N.

Fap-Fp-Fs = M*a.
240-447.4 + Fs = M*0 = 0
Fs = 207.4 N. = Force of static friction.

u = Fs/Fn = 207.4/663.2 = 0.313.

posted by Henry

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The text passage contains a physics problem and a discussion of the solution. The problem is about a box resting on an incline, and the goal is to find the coefficient of static friction between the box and the incline.
single_question_with_answer

the passage starts with the problem statement, which includes the angle of the incline (34 degrees), the parallel force required to prevent the box from sliding (240 n), and the weight of the box (800 n). it then provides a step-by-step solution, including the final answer for the coefficient of static friction (0.313).
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# Physics!

HELP!!!

A box rest on an incline making a 34 angle with the horizontal. It is found that a parallel force to the incline of at least 240 N can prevent the box from sliding down the incline. If the weight of the box is 800 N, find the coefficient of static friction between the box and the incline.

1. weight component down = m g cos 34
friction force up = m g cos 34 * mu
so
mu m g cos 34 + 240 = m g sin 34

posted by Damon
2. I get 8.01. That doesn't seem correct. I plugged in all my numbers. Anything you think I could have done incorrectly?

posted by Randy
3. I converted newtons to kilograms and my answer makes more sense to me. I got 0.312 for mu this time. Thank you for the guidance.

posted by Randy
4. M*g = 800 N.

Fp = 800*sin34 = 447.4 N.

Fn = 800*Cos34 = 663.2 N.

Fap-Fp-Fs = M*a.
240-447.4 + Fs = M*0 = 0
Fs = 207.4 N. = Force of static friction.

u = Fs/Fn = 207.4/663.2 = 0.313.

posted by Henry

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A 25 kg box slides, from rest, down a 9-m-long incline making an angle 15 degrees with the horizontal. The speed of the box when it reaches the bottom of the incline is 2.40 m/s. (a) What is the coefficient of kinetic friction
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user
The text passage contains a physics problem and a discussion of the solution. The problem is about a box resting on an incline, and the goal is to find the coefficient of static friction between the box and the incline.

TYPE: SINGLE_QUESTION_WITH_ANSWER

The passage starts with the problem statement, which includes the angle of the incline (34 degrees), the parallel force required to prevent the box from sliding (240 N), and the weight of the box (800 N). It then provides a step-by-step solution, including the final answer for the coefficient of static friction (0.313).