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2 years ago

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A 150-N box is being pulled horizontally in a wagon accelerating uniformly at 3.00 m/s2. The box does not move relative to the w

agon, the coefficient of static friction between the box and the wagon's surface is 0.600, and the coefficient of kinetic friction is 0.400. The friction force on this box is closest to
A) 450 N.
B) 90.0 N.
C) 60.0 N.
D) 45.9 N

1 answer:

Zepler [3.9K]2 years ago

8 0

Answer:

Frictional force, F = 45.9 N

Explanation:

It is given that,

Weight of the box, W = 150 N

Acceleration, $a=3\ m/s^2$

The coefficient of static friction between the box and the wagon's surface is 0.6 and the coefficient of kinetic friction is 0.4.

It is mentioned that the box does not move relative to the wagon. The force of friction is equal to the applied force. Let a is the acceleration. So,

$m=\dfrac{W}{g}$

$m=\dfrac{150}{9.8}$

$m=15.3\ kg$

Frictional force is given by :

$F=ma$

$F=15.3\times 3$

F = 45.9 N

So, the friction force on this box is closest to 45.9 N. Hence, this is the required solution.

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Generally the electric force keeping the dust particle on the air equal to the force of gravity acting on the particles

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