What is energy?

The box shown on the rough ramp above is sliding up the ramp.calculate the force of friction on the box

Anna [14]

We are given a box that slides up a ramp. To determine the force of friction we will use the following relationship:

$F_f=\mu N$

Where.

$\begin{gathered} N=\text{ normal force} \\ \mu=\text{ coefficient of friction} \end{gathered}$

To determine the Normal force we will add the forces in the direction perpendicular to the ramp, we will call this direction the y-direction as shown in the following diagram:

In the diagram we have:

$\begin{gathered} m=\text{ mass}_{} \\ g=\text{ acceleration of gravity} \\ mg=\text{ weight} \\ mg_y=y-\text{component of the weight. } \end{gathered}$

Adding the forces in the y-direction we get:

$\Sigma F_y=N-mg_y$

Since there is no movement in the y-direction the sum of forces must be equal to zero:

$N-mg_y=0$

Now we solve for the normal force:

$N=mg_y$

To determine the y-component of the weight we will use the trigonometric function cosine:

$\cos 40=\frac{mg_y}{mg}$

Now we multiply both sides by "mg":

$mg\cos 40=mg_y$

Now we substitute this value in the expression for the normal force:

$N=mg\cos 40$

Now we substitute this in the expression for the friction force:

$F_f=\mu mg\cos 40$

Now we substitute the given values:

$F_f=(0.2)(10\operatorname{kg})(9.8\frac{m}{s^2})\cos 40$

Solving the operations:

$F_f=15.01N$

Therefore, the force of friction is 15.01 Newtons.

3 0

1 year ago

1.15) What distance will 650 joules of work move a box weighing 50 newtons?

son4ous [18]

Answer:

The answer is 13 however make sure if they ask for a certain measurement like meter answer it by saying 13 meters.

Explanation:

This basically turns into basic algebra if you know the formula for work. The formula for work is W=F*d

Here are the variables that you know 650J=50N*d so you need d.

All you do is divide 650J by 50N and you get a total of 13 (meters since I don't know what they want you to put it in).

6 0

3 years ago

Learning Goal:

enot [183]

Answer:

A. $U_0 = \dfrac{\epsilon_0 A V^2}{2d}$

B. $U_1 = \dfrac{\epsilon_0 A V^2}{6d}$

C. $U_2 = \dfrac{K\epsilon_0 A V^2}{2d}$

Explanation:

The capacitance of a capacitor is its ability to store charges. For parallel-plate capacitors, this ability depends the material between the plates, the common plate area and the plate separation. The relationship is

$C=\dfrac{\epsilon A}{d}$

$C$ is the capacitance, $A$ is the common plate area, $d$ is the plate separation and $\epsilon$ is the permittivity of the material between the plates.

For air or free space, $\epsilon$ is $\epsilon_0$ called the permittivity of free space. In general, $\epsilon=\epsilon_r \epsilon_0$ where $\epsilon_r$ is the relative permittivity or dielectric constant of the material between the plates. It is a factor that determines the strength of the material compared to air. In fact, for air or vacuum, $\epsilon_r=1$ .

The energy stored in a capacitor is the average of the product of its charge and voltage.

$U = \dfrac{QV}{2}$

Its charge, $Q$ , is related to its capacitance by $Q=CV$ (this is the electrical definition of capacitance, a ratio of the charge to its voltage; the previous formula is the geometric definition). Substituting this in the formula for $U$ ,