Physics Chapter 3 passage question 5

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inescontreras
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Joined: Tue Jan 15, 2019 4:09 pm

Physics Chapter 3 passage question 5

Post by inescontreras » Tue Apr 09, 2019 9:51 pm

Could you explain how they develop that answer a bit more? I am a bit confused on how they got to the answer
NS_Tutor_Will
Posts: 538
Joined: Fri May 25, 2018 9:15 am

Re: Physics Chapter 3 passage question 5

Post by NS_Tutor_Will » Wed Apr 10, 2019 8:25 am

Thanks for the question!

The work done by a force can be expressed using this formula:

Work = Fdcosθ

We'll also need the frictional force formula, which relates the coefficient of friction to the normal force:

Ffric = μkFN

On a ramp problem like this, the normal force is equal to mgcosθ.

This means that the kinetic frictional force is equal to μkmgcosθ. We'll take m to be either 30 kg or 40 kg (but 30 ends up being right because of the answer choices).

Now we need to find the d in the Fdcosθ expression. The easiest way is to use the pythagorean theorem, but you could also use trig functions. √(8^2 + 14^2) = roughly 16 meters.

Now we're left with (μk)(m)(g)(cos30)(d)

Which equals (0.2)(30)(9.8)(cos30)(16)

I hope this helps!
inescontreras
Posts: 4
Joined: Tue Jan 15, 2019 4:09 pm

Re: Physics Chapter 3 passage question 5

Post by inescontreras » Thu Apr 11, 2019 10:42 pm

NS_Tutor_Will wrote:
Wed Apr 10, 2019 8:25 am
Thanks for the question!

The work done by a force can be expressed using this formula:

Work = Fdcosθ

We'll also need the frictional force formula, which relates the coefficient of friction to the normal force:

Ffric = μkFN

On a ramp problem like this, the normal force is equal to mgcosθ.

This means that the kinetic frictional force is equal to μkmgcosθ. We'll take m to be either 30 kg or 40 kg (but 30 ends up being right because of the answer choices).

Now we need to find the d in the Fdcosθ expression. The easiest way is to use the pythagorean theorem, but you could also use trig functions. √(8^2 + 14^2) = roughly 16 meters.

Now we're left with (μk)(m)(g)(cos30)(d)

Which equals (0.2)(30)(9.8)(cos30)(16)

I hope this helps!


Hey Will, so my issue is when the formula is plug into work (W=Fdcosθ) should you have cosθ twice? canceling each other? because just for F we will plug μkmgcosθ
giving

W = (μkmgcosθ)(d)(cosθ)
NS_Tutor_Will
Posts: 538
Joined: Fri May 25, 2018 9:15 am

Re: Physics Chapter 3 passage question 5

Post by NS_Tutor_Will » Sun Apr 14, 2019 4:38 pm

Thanks for the follow-up!

Looks like I neglected to to break down the normal force fully, so I'll try to explain. With these type of ramp problems, the normal force is the "vertical" component of gravity (perpendicular to the ramp). You can break the components down to determine where the values came from (or memorize the expression!): Fnormal = mg * cos(alpha). Alpha is the angle between the normal and the slope which is, by similar triangles, the same angle between the ground and the ramp (this makes our job easier!).

So,

Fnormal = mg cos(θ)

We also know that the work done by friction will be parallel to the surface of the ramp (there's no vertical component to friction). So now we can write our work expression:

Work = Fdcosθ, but θ is equal to 0° and cos0° is equal to 1.

So,

Work = Fd*1 = Fd


From trig, d = 8/sin 30° = 8 / 0.5 = 16 m

Normal force, from before, is mg cos(θ)

And to get back to the frictional force expression, Ffric = μkFnormal, we just need to add in the coefficient of kinetic friction: μk * Fnormal

Now we're left with (μk)(m)(g)(cosθ)(d)

Which equals (0.2)(30)(9.8)(cos30)(16)

I hope this helps clarify it!
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