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lesantik [10]
9 months ago
7

hi, Need help with triangle law of vector addition worksheet and a verifying Newtons second law worksheet?

Physics
1 answer:
Yuri [45]9 months ago
5 0

Use Newton's second law and the free body diagram to determine the net force and acceleration of an object. In this unit, the forces acting on the object were always directed in one dimension.

The object may have been subjected to both horizontal and vertical forces but there was no single force directed both horizontally and vertically. Moreover, when free-body diagram analysis was performed, the net force was either horizontal or vertical, never both horizontal and vertical.

Times have changed and we are ready for situations involving two-dimensional forces. In this unit, we explore the effects of forces acting at an angle to the horizontal. This makes the force act in two dimensions, horizontal and vertical. In such situations, as always in situations involving one-dimensional network forces, Newton's second law applies.

Learn more about Newton's second law here:-brainly.com/question/25545050

#SPJ9

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Suppose the ski patrol lowers a rescue sled carrying an injured skier, with a combined mass of 97.5 kg, down a 60.0-degree slope
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a. 1337.3 J work, in joules, is done by friction as the sled moved 28 m along the hill.

b.21,835 J work, in joules, is done by the rope on the sled this distance.

c. 23,170 J   the work, in joules done by the gravitational force on the sled d. The net work done on the sled, in joules is 43,670 J.

       

<h3>What is friction work?</h3>

The work done by friction is the force of friction times the distance traveled times the cosine of the angle between the friction force and displacement

a. How much work is done by friction as the sled moves 28m along the hill?

ans. We use the formula:

friction work = -µ.mg.dcosθ

  = -0.100 * 97.5 kg * 9.8 m/s² * 28 m * cos 60

= -1337.3 J

-1337.3 J work, in joules, is done by friction as the sled moved 28 m along the hill.

b. How much work is done by the rope on the sled in this distance?

We use the formula:

Rope work = -m.g.d(sinθ - µcosθ)

rope work = - 97.5 kg * 9.8 m/s² * 28 m (sin 60 – 0.100 * cos 60)

                     = 26,754 (0.816)

                     = 21,835 J

21,835 J work, in joules, is done by the rope on the sled this distance.

c.  What is the work done by the gravitational force on the sled?

By using  the formula:

Gravity work = mgdsinθ

                    = 97.5 kg * 9.8 m/s² * 28 m * sin 60

                    = 23,170 J

23,170 J   the work, in joules done by the gravitational force on the sled .

       

D. What is the total work done?

By adding all the values

work done =  -1337.3 + 21,835 + 23,170

                 = 43,670 J

The net work done on the sled, in joules is 43,670 J.

Learn more about friction work here:

brainly.com/question/14619763

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