Question

A 15.0 m long plane is inclined at 30.0 degrees. If the coefficient of friction is
0.426, what force is required to move a 40.0 kg mass from rest at the bottom on
the plane to the top of the plane with a final velocity of 8.00 m/s?
How would I set up this problem?

Answer 1

Hi,

Do you know how to draw the free body diagram of the situation described?

Answer 2

@Joq Yes, but I'm not sure if I drew the diagram correctly.

Answer 3

@Jamie_S Can you post it here? Maybe it's fine.

Answer 4

@Joq #3 is for this problem

Answer 5

Let me see.

Answer 6

Did it post?

Answer 7

Yep. I'm looking at it.

Answer 8

The diagram is good. Suppose you put the box in the incline an you are not exerting any upward force.

Answer 9

It will move downwards, right? That's because it has a component of its weight parallel to the slope.

Answer 10

If you want to move it upward, you'll have to overcome that chunk of its weight and the friction that is always opposite to the motion.

Answer 11

Wait...it will move downwards? The problem indicates it starts from rest at the bottom and is pushed/pulled to the top.

Answer 12

Oh, okay. How would I find the force needed to result in a velocity of 8.0m/s?

Answer 13

A \(F_{net}\) brings the box from the bottom with \(v_0 =0\) to the top with \(v=8.00\ m/s\) through a distance of \(15\ m\).

Answer 14

We know the mass but the acceleration do not. We need to figure that first. Do you know an equation that relates distance, initial velocity, final velocity and acceleration?

Answer 15

Would the correct equation be 2ad=v^2 ?

Answer 16

Yep. Solving for a.

To make sure what we are doing here: We know that a component of the weight and friction oppose the upward force exerted by the person. This is:

\(\large{F_{net}= F_{exerted} - friction - component\ of \ weight}\)

If we find \(F_{net}\) we can solve for \(F_{exerted} \) and we're done.

Answer 17

Now we are finding \(F_{net}\). The equation you've posted will help us to do that.

Answer 18

Hmm, would the equation be a Fnet = Fexerted - 0.426 - 40.0kg?

Answer 19

Nope. If you want to know what the component of the weight is, you can look at the drawing you've drawn.

Answer 20

The weight goes straigt down. And it's the hypotenuse of that triangle. The side parallel to the slope, \(w_{||}\), (the component we're interested in) is the opposite side of the angle we know of: 30º (do you have a doubt in this point?).

So:

\(\large{\sin(30º)= \frac{w_{||}}{w}}\)

\(\large{w\sin(30º)= w_{||}}\) <- Multiply both sides by \(w\).

Answer 21

That is the component of the weight opposing the upward motion.

Answer 22

I'm completely lost. And it has to do with the way the site formatted your response (there are literally 15 "?" marks after the number 30).

Answer 23

Actually more...

Answer 24

Aps, refresh the browser.

Answer 25

After doing so it should be solved that problem.

Answer 26

Ahh, okay I see now. Should there be a number replacing w?

Answer 27

Yep. w=mg.

m= mass.
g= acceleration due to gravity.

Answer 28

So, mass would be 40kg and acceleration would be 2m/s^2? I did a = (8.0m/s)^2/(2)(15.0m)

Answer 29

Yep.

But notice that you're given values to 3 significant figures, so the acceleration you've found should be to 3 significant figure as well.

Answer 30

2.13 m/s^2

Answer 31

Aps, but don't confuse the acceleration due to gravity with the acceleration when going upward.

Answer 32

Acceleration due to gravity is used to find its weight, w and the other is to find the Fnet.

Answer 33

The equation now is:

\(\large{F_{net}= F_{exerted}-friction- w\sin(\theta)}\)

\(\large{F_{net}= F_{exerted}-[friction+w\sin(\theta)]}\)

\(\large{F_{exerted}= F_{net}+friction+ w\sin(\theta)}\)

Answer 34

We need to find friction too, but find Fnet and wsin(30º) first.

Answer 35

is w 48kg * 2.13m/s^2 ?

Answer 36

Yep. 40.0kg * 2.13 m/s^2 is the Fnet.

Answer 37

Now the component of the force.

Answer 38

Would Fnet = 181.476m/s^2?

Answer 39

If an acceleration of 2.13m/s^2 is applied to a 40.0kg mass, then the Fnet is:

\(F_{net}=ma = (40.0\ kg)(2.13\ m/s^2) = 85.3\ N \)

Answer 40

85.3 N instead of 85.2 N because I didn't round the acceleration, since it is not the final result that is being asked to find.

Answer 41

I put (2.13m/s)^2 instead...oops

Answer 42

So Fnet = 85.3N ... Now would the next equation be 85.3Nsin(30 degrees)?

Answer 43

Nope. It's much easier to solve each term of the equation independently first, and then bring all together.

Answer 44

You've found Fnet. Now you need to find wsin(30 degrees).

Answer 45

Once done so, we need to find friction.

Answer 46

Is w 2.13m/s^2?

Answer 47

wsin(30 degrees)

w is the weight. And the weight on the Earth is the acceleration g applied to an object of mass m: mg. Do you know what g is?

Answer 48

9.8?

Answer 49

Yep.

Answer 50

I forget the units. Is it 9.8m/s^2?

Answer 51

Yep. Or 9.80 m/s^2 if we want to be consistent with the number of significant figures of the values given at the beginning.

Answer 52

Would it be 4.90m/s^2?

Answer 53

Yep. That would be the component of the acceleration of gravity acting parallel to the slope. That's acting on the mass m.

Answer 54

Okay what's next?

Answer 55

Once you find the value of wsin(30 degrees), friction needs to be found.

Do you know how friction is defined? What is its formula?

Answer 56

Ffr = uFn?

Answer 57

What Fn is?

Answer 58

I'm not sure to be honest

Answer 59

That is the Normal Force. The Normal Force is always perpendicular to the surface where an object is supported.



If you are sat now without and there is no force moving you downward or upward, then the Normal Force is equal to your weight. In an incline, the Normal Force looks like this: