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anygoal [31]
3 years ago
12

A student pulls on a 20 kg box with a force of 50 N at an angle 45 degrees relative to the horizontal. The box increases

Physics
1 answer:
telo118 [61]3 years ago
8 0

The value of the friction force is 5.4 N

Explanation:

In order to solve the problem, we have to write down the equation of the forces along the horizontal direction.

We have only two forces acting along the horizontal direction:

  • The horizontal component of the pull of the student, F cos \theta, forward
  • The friction force, F_f, backward

The resultant force must be equal to the product between mass and acceleration (Newton's second law), so we can write:

F cos \theta - F_f = ma

where:

F = 50 N is the force of pull of the student

\theta=45^{\circ} is the angle relative to the horizontal

m = 20 kg is the mass of the box

a=1.5 m/s^2 is the acceleration

Solving the equation for F_f, we find the value of the friction force:

F_f = Fcos \theta -ma=(50)(cos 45^{\circ})-(20)(1.5)=5.4 N

Learn more about friction:

brainly.com/question/6217246

brainly.com/question/5884009

brainly.com/question/3017271

brainly.com/question/2235246

#LearnwithBrainly

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Answer:

0.208 N

Explanation:

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F_1=F_3=F

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3 0
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normal force on the ball = ?          

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normal force acting on the ball will be

F n = F - mg                          

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Answer:

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work done = F s cos 90°        ∵ cos 90° = 0

Work done = 0 J

8 0
3 years ago
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