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A 4.04 kg block slides down a smooth, frictionless plane having an inclination of 30◦. The acceleration of gravity is 9.8 m/s^2. The acceleration of the plane is 4.9
What is the magnitude of the perpendicular force that the block exerts on the surface of the plane at a distance of 2.37 m down the incline?
Answer in units of N.

Sagot :

Luv2Teach

Answer:

Explanation:

What a lot of words to solve such a simple problem! The perpendicular force is the one that is pushing straight down on the plane. There is no side to side movement here or friction acting on this dimension at all. Perpendicular force is the same as the weight of the block. That's it! Perpendicular force is also normal force which is the same thing as weight:

w = mg so

w = (4.04)(9.8) and

w = 4.0 × 10¹ N

temdan2001

The magnitude of the perpendicular force that the block exerts on the surface of the plane at a distance of 2.37 m down the incline plane is 34.3N

The given parameter are:

mass M = 4.04 kg

Angle of inclination = 30 degree

Acceleration due to gravity = 9.8 m/s^s

To calculate the magnitude of the perpendicular force that the block exerts on the surface of the plane at a distance of 2.37 m down the plane, The force will be the weight of the block which is equal to the normal reaction.

Normal N = mgcosФ

N = 4.04 x 9.8 x cos30

N = 34.28 Newtons

At all point in the plane,  the magnitude of the perpendicular force that the block exerts on the surface of the plane will be the same.

Therefore,  the magnitude of the perpendicular force that the block exerts on the surface of the plane at a distance of 2.37 m down the incline plane is 34.3N

Learn more here : https://brainly.com/question/23944424

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