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3 years ago

A 7kg block slides down a 30 degree slope at a constant velocity of 2 m/s. How big is the force of friction acting on the block?

Physics

1 answer:

lisabon 2012 [21]3 years ago

6 0

The force of friction is <u>34.3 N.</u>

A block of mass m slides down a plane inclined at an angle θ to the horizontal with a constant velocity. According to Newton's first law of motion, every body continues in its state of rest or a state of uniform motion in a straight line, unless acted upon, by an external unbalanced force. This means that when balanced forces act on a body, the body moves with a constant velocity.

The free body diagram of the sliding block is shown in the attached diagram. Resolve the weight mg of the block into two components mg sinθ along the direction of the plane and mg cosθ perpendicular to the plane . The force of friction F acts upwards along the plane and the normal reaction acts perpendicular to the plane.

Since the block moves down with a constant velocity, the downward force mg sinθ must be equal to the upward frictional force.

$F = mg sin\theta$

Substitute 7 kg for m, 9.8 m/s² for g and 30° for θ.

$F = mg sin\theta = (7 kg)(9.8 m/s^2)(sin30^o) =34.3 N$

The force of friction is <u>34.3 N</u> up the plane.

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we substitute

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Therefore, magnitude is 1382.59 N/C

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