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Ludmilka [50]
2 years ago
9

Peter does 3,000 joules (J) of work by moving an object 40 meters (m). What is the force that Peter applies to the object? A 75

N B 2,960 N C 3,040 N D 120,000 N
Physics
1 answer:
Papessa [141]2 years ago
4 0

The force that peter applies to the object of distance 40m is 75N.

<h3>HOW TO CALCULATE FORCE</h3>

The force applied to an object can be calculated by dividing the work done on the object by the distance moved. That is;

Force = Work done ÷ distance

According to this question, the work done is 3000 joules while the distance moved is 40m. The force is calculated as follows:

Force = 3000J ÷ 40m

Force = 75N

Therefore, the force that peter applies to the object of distance 40m is 75N.

Learn more about force at: brainly.com/question/26115859

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A 50.0-kg block is being pulled up a 16.0° slope by a force of 250 n that is parallel to the slope. the coefficient of kinetic
Drupady [299]
When dealing with multiple forces acting on a body, it is advisable to draw a free-body diagram like that shown in the picture. There are four forces acting on the box: weight (W) pointing straight down, normal force perpendicular to the slope denoted as Fn, force used to push the box upwards along the slope and the frictional force acting opposite to the direction of motion of the box denoted as Ff. Frictional force is equal to coefficient of kinetic friction (μk) multiplied with Fn.

∑Fy = Fn - mgcos30° = 0
           Fn = (50)(9.81)(cos 16) = 471.5 N

When in motion, the net force is equal to mass times acceleration according to Newton's 2nd Law of Motion:

Fnet = F - μk*Fn - mgsin30° = ma
          250 - (0.2)(471.5 N) - (50)(sin 16°) = (50)(a)
          a = 2.84 m/s²

8 0
3 years ago
Take a look at a bicycle with gears. Using what you have just learned, answer the following questions:
Harlamova29_29 [7]

Answer:

answer a: a large front gear with a small back gear

answer b: a small front gear with a large back gear

Explanation:

just simple gearing ratios

4 0
2 years ago
An object in free fall is dropped from a building. Its starting velocity is 0 m/s. Ignoring the effects of air resistance, what
Amanda [17]

Answer:

30m/s^2

Explanation:

Acceleration=Final Velocity-Initial Velocity/Time

10m/s^2= Final Velocity-0m/s/3

30m/s^2= Final Velocity

Final Velocity=30m/s^2

7 0
2 years ago
A 5.50 kg sled is initially at rest on a frictionless horizontal road. The sled is pulled a distance of 3.20 m by a force of 25.
kiruha [24]

(a) 69.3 J

The work done by the applied force is given by:

W=Fd cos \theta

where:

F = 25.0 N is the magnitude of the applied force

d = 3.20 m is the displacement of the sled

\theta=30^{\circ} is the angle between the direction of the force and the displacement of the sled

Substituting numbers into the formula, we find

W=(25.0 N)(3.20 m)(cos 30^{\circ})=69.3 J

(b) 0

The problem says that the surface is frictionless: this means that no friction is acting on the sled, therefore the energy dissipated by friction must be zero.

(c) 69.3 J

According to the work-energy theorem, the work done by the applied force is equal to the change in kinetic energy of the sled:

\Delta K = W

where

\Delta K is the change in kinetic energy

W is the work done

Since we already calculated W in part (a):

W = 69.3 J

We therefore know that the change in kinetic energy of the sled is equal to this value:

\Delta K=69.3 J

(d) 4.9 m/s

The change in kinetic energy of the sled can be rewritten as:

\Delta K=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2 (1)

where

Kf is the final kinetic energy

Ki is the initial kinetic energy

m = 5.50 kg is the mass of the sled

u = 0 is the initial speed of the sled

v = ? is the final speed of the sled

We can calculate the variation of kinetic energy of the sled, \Delta K, after it has travelled for d=3 m. Using the work-energy theorem again, we find

\Delta K= W = Fd cos \theta =(25.0 N)(3.0 m)(cos 30^{\circ})=65.0 J

And substituting into (1) and re-arrangin the equation, we find

v=\sqrt{\frac{2 \Delta K}{m}}=\sqrt{\frac{2(65.0 J)}{5.50 kg}}=4.9 m/s

6 0
3 years ago
Your friend provides a solution to the following problem. Evaluate her solution. The problem: Jim (mass 50 kg) steps off a ledge
Mekhanik [1.2K]

Answer:

No its wrong

Correct compression is 0.41

Explanation:

After jumping from 2m height on spring attached platform platform is compressed a distance x(let).

So work done by gravity on Jim is converted into spring potential energy.

k=8000 N/m

mass of Jim =50 kg

\frac{1}{2}k(x)^{2} =m*g(2+x)

\frac{1}{2}8000(x)^{2} =50*9.8(2+x)

8.16x^{2} =2 +x

Solve this quadratic one solution is positive and other is negative.

positive one is our answer = 0.41 m

5 0
3 years ago
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