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zmey [24]
3 years ago
8

Pls help :/

Physics
1 answer:
vodomira [7]3 years ago
6 0

Answer:

Gravitational attraction of the sun.

Explanation:

Gravity is an attractive force. Any two masses will exert an attractive force on the other according to Newton's law of universal gravitation. The more massive the objects, the stronger the force. The sun, as you can probably guess, is pretty massive - 330,000 times more than Earth, and 1,048 time more than Jupiter, our solar system's largest planet. Just like man-made satellites around Earth, the planets in our solar system are constant process of "falling" around the sun, locked in their orbits by its mass, but slowing dramatically in their orbital velocity the further away they are.

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Mrac [35]

The general formula to calculate the work is:

W=Fd \cos \theta

where F is the force, d is the displacement of the couch, and \theta is the angle between the direction of the force and the displacement. Let's apply this formula to the different parts of the problem.


(a) Work done by you: in this case, the force applied is parallel to the displacement of the couch, so \theta=0^{\circ} and \cos \theta=1, therefore the work is just equal to the product between the horizontal force you apply to push the couch and the distance the couch has been moved:

W=Fd=(220.0 N)(3.9 m)=858 J


(b) work done by the frictional force: the frictional force has opposite direction to the displacement, therefore \theta=180^{\circ} and \cos \theta=-1. Therefore, we must include a negative sign when we calculate the work done by the frictional force:

W=-Fd=-(144.0 N)(3.9 m)=-561.6 J


(c) The work done by gravity is zero. In fact, gravity (which points downwards) is perpendicular to the displacement of the couch (which is horizontal), therefore \theta=90^{\circ} and \cos \theta=0: this means

W=0.


(d) Work done by the net force:

The net force is the difference between the horizontal force applied by you and the frictional force:

F=220 N-144 N=76 N

And the net force is in the same direction of the displacement, so \theta=0^{\circ} and \cos \theta=1 and the work done is

W=Fd=(76 N)(3.9 m)=296.4 J


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hope this helped!

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