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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

A car traveling at 23m/s starts to decelerate steadily. It comes to a complete stop in 7 seconds. What is the acceleration?

Lena [83]

The car has an initial velocity $v_0$ of 23 m/s and a final velocity $v$ of 0 m/s. Recall that for constant acceleration,

$v=v_0+at$

The car stops in 7 s, so the acceleration is

$0\,\dfrac{\mathrm m}{\mathrm s}=23\,\dfrac{\mathrm m}{\mathrm s}+a(7\,\mathrm s)$

$\implies a\approx-3.29\,\dfrac{\mathrm m}{\mathrm s^2}$

7 0

3 years ago

O operate this generator, a human applies a force by turning the crank. This action generates an electric current, which lights

Shtirlitz [24]

Answer A I’m very positive about it

3 0

3 years ago

Read 2 more answers

A sled slides along a horizontal surface on which the coefficient of kinetic friction is 0.25. Its velocity at point A is 8.4 m/

Bad White [126]

Answer:

1.06 secs

Explanation:

Initial speed of sled, u = 8.4 m/s

Final speed of sled, v = 5.8 m/s

Coefficient of kinetic friction, μ = 0.25

Using the impulse momentum theory, we know that the impulse applied to the sled is equal to change in momentum of the sled:

FΔt = mv - mu

where m = mass of the object

Δt = time interval

F = force applied

The force applied on the sled is the frictional force, which is given as:

F = -μmg

where g = acceleration due to gravity

Therefore:

-μmgΔt = mv - mu

-μmgΔt = m(v - u)

-μgΔt = v - u

Making Δt subject of formula:

Δt = (v - u) / -μg

Δt = (5.8 - 8.4) / (-0.25 * 9.8)

Δt = -2.6/ -2.45

Δt = 1.06 secs

It took the sled 1.06 secs to travel from A to B.

7 0

3 years ago

A type of Lead (Pb) ion has a + 4 oxidation number. Sulfur (S) has a — 2 oxidation number. What would be the chemical formula fo

serious [3.7K]

Answer: The chemical formula for the compound of these two elements is $PbS_2$

Explanation:

For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.

Here metal lead is having an oxidation state of +4 called as $Pb^{4+}$ cation and sulphur non metal has oxidation state of -2 called as $S^{2-}$ . Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral $PbS_2$

The chemical formula for the compound of these two elements is $PbS_2$

4 0

3 years ago