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Nuetrik [128]
3 years ago
12

A book weighing 5 N rests on top of a table. 1) A downward force of magnitude 5 N is exerted on the book by the force of

Physics
1 answer:
andreev551 [17]3 years ago
7 0

Answer:

1) A downward force of magnitude 5 N is exerted on the book by the force of of gravity

2) An upward force of magnitude 5 N is exerted on the book by the table

Explanation:

First of all, any object near the Earth's surface experiences the forces of gravity, which is also called weight of the object. This force always acts downward.

For the book in the problem, the magnitude of the weight is 5 N.

We also know that the book is at rest: this means that the net force acting on it is zero, and there must be another force balancing the weight, in order to give a zero net force. This other force is the reaction force exerted by the table on the book: the magnitude of this force must be equal to the force of gravity (so, 5 N) and its direction is opposite to the weight, therefore upward.

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7. Two people are pushing a 40.0kg table across the floor. Person 1 pushes with a force of 490N
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Answer:

20.4 m/s^{2}

Explanation:

To start doing this problem, first draw a free body diagram of the table. My teacher always tells us to do this, and I find that it is very helpful. I have attached a free body diagram to this answer- take a look at it.

First, let us see if Net force = MA. To do that, we need to determine whether the object is at equilibrium horizontally. For an object to be at equilibrium, it either needs to be moving at a constant velocity or not moving at all. Also, if an object is at equilibrium, there will not be any acceleration. But we know that there IS acceleration horizontally, so it cannot be in equilibrium. If it is not in equilibrium, we can use the formula ∑F= ma.

Let us determine the net force. Since the object is moving horizontally, we can ignore the weight and normal force, because they are vertical forces. The only horizontal forces we need to worry about are the applied force and force of friction.

Applied force = 1055 N (490 + 565)

Friction force= Unknown

To find the friction force, use the kinetic friction formula, Friction = μkN

μk is the coefficient, which the problem includes- it is 0.613.

N is the normal force, which we have to find.

*To find the normal force, we have to determine if the object is at equilibrium VERTICALLY. Since it has no acceleration vertically (it's not moving up/down), it is at equilibrium. Now, when an object is at equilibrium in one direction, it means that all the forces in that direction are equal. What are our vertical forces? Weight (mg) and Normal force (N). So it means that the Normal force is equal to the Weight.

Weight = mg = (40)(9.8) = 392 N

Normal force = 392 N

Now, plug it back into the formula (μkN): (0.613)(392) = 240.296 N

Friction = 240.296 N

Now that we know the friction, we can find the horizontal net force. Just subtract the friction force, 240.296 from the applied force, 1055 N

Horizontal Net Force: 814.704 N

Now that we know the net force, plug in the numbers for the formula

∑F= ma.

814.704 = (40.0)(a)

*Divide on both sides)

a = 20.3676 m/s^2

Round it to 3 significant figures, to get:

20.4 m/s^{2}

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