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#forces
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#mechanics
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#physics
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A mass of 10 tonnes is being hauled along a level track and at a certain instant the velocity is 9km/h and the acceleration 0.2m/s square. If the total friction is 1200 find the pull in the rope?

Mar 9

And the power exerted at the instant?

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Mar 9

To find the pull in the rope, we need to consider the forces acting on the mass.

The forces acting on the mass are the force of gravity (weight), the force of friction, and the force of tension in the rope.

1. Force of gravity (weight) = mass x acceleration due to gravity

Fg = 10,000 kg x 9.8 m/s^2 = 98,000 N

2. Force of friction = 1200 N

3. Net force = mass x acceleration

Fnet = 10,000 kg x 0.2 m/s^2 = 2000 N

Now, we can find the force of tension in the rope by using Newton's second law:

Ft - Ff - Fg = Fnet

Ft - 1200 N - 98000 N = 2000 N

Ft = 2000 N + 1200 N + 98000 N

Ft = 100,200 N

Therefore, the pull in the rope is 100,200 N.

The forces acting on the mass are the force of gravity (weight), the force of friction, and the force of tension in the rope.

1. Force of gravity (weight) = mass x acceleration due to gravity

Fg = 10,000 kg x 9.8 m/s^2 = 98,000 N

2. Force of friction = 1200 N

3. Net force = mass x acceleration

Fnet = 10,000 kg x 0.2 m/s^2 = 2000 N

Now, we can find the force of tension in the rope by using Newton's second law:

Ft - Ff - Fg = Fnet

Ft - 1200 N - 98000 N = 2000 N

Ft = 2000 N + 1200 N + 98000 N

Ft = 100,200 N

Therefore, the pull in the rope is 100,200 N.