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Tcecarenko [31]
3 years ago
15

A 400-kg object has a 100-Newton rightward net force being applied to it. What is the magnitude of the rightward acceleration on

the object (in m/s^2)?
Physics
1 answer:
aliya0001 [1]3 years ago
5 0

Answer:

The answer to your question is a = 0.25 m/s²

Explanation:

Data

mass = m = 400 kg

Force = F = 100 N

acceleration = a = ? m/s²

Process

To solve this problem use Newton's second law that states that the force applied to an object is directly proportional to the mass of the body times its acceleration.

Formula

                       F = ma

solve for a

                       a = \frac{F}{m}

Substitution

                       a = \frac{100}{400}

Simplification and result

                              a = 0.25 m/s²

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

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<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

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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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