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Rzqust [24]
3 years ago
10

A student applied a force of 200 N to push a box 15 m down the sidewalk. Given the formula, W=F×d, how much work has been done?

Physics
1 answer:
Ede4ka [16]3 years ago
7 0

Answer: 3,000

Explanation: 200 times 15

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Acceleration is defined as the rate of change for which of the following? Question 4 options: time position velocity displacemen
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Acceleration can be defined as the rate of change in the velocity of an object. Option C is correct.

<h3>What is Acceleration?</h3>
  • It is defined as the rate of change in velocity.
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a =  \dfrac {v-u}t\ \ \ \ \rmor}\\\\a = \dfrac {\Delta v }t

Where,

a - acceleration

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t - time

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Carbon has six protons. which model shows a neutral atom of carbon?
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Jake is helping Fin push a box at a constant velocity up an incline that makes an angle of 30.0° above the horizontal by applyin
andre [41]

Given data

The angle of inclination of the plane is theta = 30 degree

The applied force in the inclined plane is F = 94 N

The distance moved in the inclined plane is d = 2.30 m

The coefficient of kinetic friction is u_k = 0.280

The free-body diagram of the above configuration is shown below:

Here, the normal reaction force on the box is N, the acceleration due to gravity is denoted as g, the friction force on the box is F_f, and the mass of the box is denoted as m.

(a)

The expression for the work done by the pushing force is given as:

W=Fd

Substitute the value in the above equation.

\begin{gathered} W=94\text{ N}\times2.30\text{ m} \\ W=216.2\text{ J} \end{gathered}

Thus, the work done by the pushing force is 216.2 J.

(b)

The box is moving at the constant velocity, therefore, the pushing force will be equal to the frictional force and the component of the gravitational force in the inclined plane.

\begin{gathered} F=F_f+mg\sin \theta \\ F=\mu_kN+mg\sin \theta \end{gathered}

The expression for the normal reaction force is given as:

N=mg\cos \theta

The expression for the mass of the box is given as:

\begin{gathered} F=\mu_k\times mg\cos \theta+mg\sin \theta \\ m=\frac{F}{\mu_kg\cos \theta+g\sin \theta} \end{gathered}

Substitute the value in the above equation.

\begin{gathered} m=\frac{94\text{ N}}{0.28\times9.8m/s^2\times\cos 30^o+9.8m/s^2\times\sin 30^0} \\ m=12.9\text{ kg} \end{gathered}

Thus, the mass of the box is 12.9 kg.

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1 year ago
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