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tensa zangetsu [6.8K]
3 years ago
11

A high school physics teacher also happens to be the junior hockey team coach. During a break at practice, the coach asks two pl

ayers to go to the center of the ice with a 10.0-m pole. A 40-kg player is at one end of the pole and a 60-kg player is at the other end. The players then start pulling themselves together by pulling the rod and sliding on the ice as they move along the rod. When the two players meet, what distance will the 60-kg player have moved
Physics
1 answer:
yawa3891 [41]3 years ago
4 0

Answer:

Explanation:

Since no external force acts on the system , centre of mass of two players will remain the same throughout the movement .

Distance of centre of mass is inversely proportional to mass . So centre of mass of the system will be nearer to 60 kg player

Distance of cetre of mass from 60 kg player

= 10 x 40 / (60+40 )

= 4 m

They will meet at centre of mass .

So, 60 kg player will be moving by a distance of 4 m , 40 kg player will be moving by 6 m before they meet .

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7. Two people are pushing a 40.0kg table across the floor. Person 1 pushes with a force of 490N
artcher [175]

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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