Energy Question

longwave

Hello everyone, I'm new here.

I was wondering if any of you energy experts could help me with a problem. I don't even know where to start.

I hope I've posted this thread in the right place. It looks the most appropriate but please feel free to move it if there is a more appropriate place.

Anyway, here is the question:

A particle of mass (m) is projected with speed (v) up a rough slope which is inclined at an angle (a) to the horizontal.

The coefficient of friction between the particle and the slope is u.

Show that the maximum distance (x) travelled up the slope by the particle at the point at which it's Kinetic Energy is at a minimum and it's Potential Energy is at a maximum is given by the formula:

x = v^2/(2g(sin a - u.cos a))

If a = arcsin (0.6) and u = 0.5, show that the particle will have lost 80% of it's initial Kinetic Energy by the time it slides back to the bottom of the slope.


Thank you in advance for any assistance anyone can offer

longwave

Anyone? Please.

societys_child

This board was called "Utilities" until it was recently renamed "Energy" . . . now look whats happened :rotfl:

I hope I've posted this thread in the right place.

Sorry OP, I really don't know. Normally discussions on here are about the price of gas and electricity . . . 'n' stuff like that.

Richie-from-the-Boro

societys_child wrote: »

This board was called "Utilities" until it was recently renamed "Energy" . . . now look whats happened :rotfl:




Sorry OP, I really don't know. Normally discussions on here are about the price of gas and electricity . . . 'n' stuff like that.

- agreed
- first of many thousands

"forum design can have unforeseen consequences, I for one don't want to wade through or filter totally irrelevant subtribe posts to find what I want"

penrhyn

There is a chap at Woolsthorpe Manor who might know, sadley I only took A level Physics for one year.

cmooney

Hi,

This is clearly an A-level or GCSE homework, but being a self described scientist it did my nut that I didn't come up with the solution straight away.

I'm only going to answer the first part, the derivation, because this is where the skill lies and the second part is just plugging numbers in.

The derivation consists of two parts also you have a total distance x, that has to be traveled but two opposite forces, gravity and friction. We shall calculated their contributions and called the y and w.

x = y - w

First derive an expression to a mass without friction then with.

The maximum distance traveled is when the potential energy of the mass is equal to it's kinetic energy.

1/2 * m * v^2 = mgh

rearrange for h is the maximum distance travelled.

v^2/2g = h

However this is only true for something being thrown in the air. Your example includes an incline using trigonometry we can replaced h with an equivalent distance along an incline,

h = y sin a

so,

v^2 /2g = y sin a, rearranged is,

v^2/2g sin a = y

Now we have to include your friction coefficient, u

Again we start from a simpler example of a mass moving along a flat plane. Here the distance traveled would be the kinetic energy against the friction work,

The work done by friction is,

W = umgd,

this would be equal to the kinetic energy,

1/2 * mv^2 = umgd

cancel and rearrange gives,

v^2 /2ug = d

Now again you are on an inclined plane and the friction is proportional to the weight in the plane. Again some trigonometry is required. So the actual effective mass is proportional to cos a, so,

v^2/2ug cos a = w

To recap we have have a total distance required to be traveled, x and it's two constituents, w and y

x = y - w

x = v^2/2g sin a - v^2/2ug cos a

Re-arrange and you get the answer you were looking for.

I haven't checked for typos, or really made any effort to make this clear as I'm sure you really shouldn't be asking for help.

I_have_spoken

I feel sick

penrhyn

I'm not worthy!

zagfles

cmooney wrote: »

Hi,

This is clearly an A-level or GCSE homework, but being a self described scientist it did my nut that I didn't come up with the solution straight away.

I'm only going to answer the first part, the derivation, because this is where the skill lies and the second part is just plugging numbers in.

The derivation consists of two parts also you have a total distance x, that has to be traveled but two opposite forces, gravity and friction. We shall calculated their contributions and called the y and w.

x = y - w

First derive an expression to a mass without friction then with.

The maximum distance traveled is when the potential energy of the mass is equal to it's kinetic energy.

1/2 * m * v^2 = mgh

rearrange for h is the maximum distance travelled.

v^2/2g = h

However this is only true for something being thrown in the air. Your example includes an incline using trigonometry we can replaced h with an equivalent distance along an incline,

h = y sin a

so,

v^2 /2g = y sin a, rearranged is,

v^2/2g sin a = y

Now we have to include your friction coefficient, u

Again we start from a simpler example of a mass moving along a flat plane. Here the distance traveled would be the kinetic energy against the friction work,

The work done by friction is,

W = umgd,

this would be equal to the kinetic energy,

1/2 * mv^2 = umgd

cancel and rearrange gives,

v^2 /2ug = d

Now again you are on an inclined plane and the friction is proportional to the weight in the plane. Again some trigonometry is required. So the actual effective mass is proportional to cos a, so,

v^2/2ug cos a = w

To recap we have have a total distance required to be traveled, x and it's two constituents, w and y

x = y - w

x = v^2/2g sin a - v^2/2ug cos a

Re-arrange and you get the answer you were looking for.

I haven't checked for typos, or really made any effort to make this clear as I'm sure you really shouldn't be asking for help.

Or an (arguably) easier way.

Gravity resolved in the direction of the slope = mg sin a
Normal force = mg cos a
Friction = Normal force * u

So force acting down the slope = gravity plus friction
= mg sin a + umg cos a = mg (sin a + u cos a )

F=ma, so acceleration down the slope divide by m so
acceleration down the slope = g (sin a + u cos a)
OR accelation up the slope = -g(sin a + u cos a)
(easier if we use up as positive)

Use the equation of motion "v^2 = u^2 + 2as" where v is the final velocity, u is initial velocity, a is acceleration and s is distance (don't confuse the letters in the formula with the letters we're using in the question!)

In this case the initial velocity is v and the final velocity is 0, ie the point when it reaches the max height. Distance is x.

So 0 = v^2 -2g(sin a + u cos a)x

and so x = v^2/2g(sin a + u cos a)

ETA: just seen the original given formula was x = v^2/(2g(sin a - u.cos a))
This can't possibly be right because x would be infinite if sin a = u cos a, ie the particle would travel up the slope forever!! OP check the question again, pretty sure my answer is right!

cmooney

Hi,

I bow out. My initial assumption of x = y-w was incorrect and doesn't apply to this problem.

I can't find fault with Zagfles' derivation.

I hand back my Physics A-level with humility.

Ciar!n

bubblesbonbon

cmooney wrote: »

Hi,

I bow out. My initial assumption of x = y-w was incorrect and doesn't apply to this problem.

I can't find fault with Zagfles' derivation.

I hand back my Physics A-level with humility.

Ciar!n

Remember - Never drink and derive