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I am Lyosha [343]
3 years ago
8

A student pushes on a 20.0 kg box with a force of 50 N at an angle of 30° below the horizontal. The box accelerates at a rate of

0.5 m/s2. What is the value of the friction force on the box?
200 N
50 N
43 N
33 N
Physics
2 answers:
anygoal [31]3 years ago
8 0
F - F(friction) = m.a
50cos(30) - F(friction) = 20 x 0.5
F(frictions) = 50cos(30) - 10 = 33.3 N
ikadub [295]3 years ago
6 0

Answer:

<u>Option-(D):</u> The value of the frictional force will be around 33 N.

Explanation:

So, we have a force, F of 50 N which is applied by a boy on a box of mass around 20 kg with an angle of 30 degree. We have the acceleration,a of 0.5 m/sec².As we know that each actions get some opposition and when we apply some force,F to an object there is a minimum amount of opposing force,F(friction) to it. And to find out the value of the frictional force,F(friction) we have the following equation to consider.

  • <u>F(applied) - Friction= m.a,</u>
  1. F×cos(angle)-friction(tot)=mass(m).acceleration(a),
  2. Now, putting all the required values inside the equation:
  3. 50×cos(30)-frictional force(F)=(20).(0.5),
  4. 50×(0.866)-frictional force(F)=10,
  5. 43.3-frictional force(F)=10,
  6. frictional force(F)=43.3-10,
  7. <u>F=33.3 N,</u>⇒Answer.

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high school physics, no need detail explain, just give the answer, but you have to make sure thank you
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Answer:

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Explanation:

If it takes the cannonball 2 seconds to reach the maximum height, we can use the analysis of the vertical component of the velocity and the fact that the acceleration of gravity is the one acting opposite to this initial vertical component (v_y) of the velocity. We know as well that at the top of the trajectory, the vertical component of the velocity is zero, and then the movement starts going down in it trajectory. So, the final velocity for the first part of the ascending movement is zero, giving us the following equation for the velocity under an accelerated movement (with acceleration of gravity "g" acting):

v_f=v_i-a\,t\\v_f=v_y-g\,t\\0=v_y-9.8\,*\,2\\v_y=9.8\,*\,2=19.6 \frac{m}{s}

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sin(\theta)=\frac{opp}{hyp} \\sin(\theta)=\frac{19.6}{40}\\\theta=arcsin(\frac{19.6}{40})\\\theta=29.34^o

which tells us that the closer answer shown is 30^o

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