Hello - basic incline mass and energy question, thanks :)

[Bix]

Homework Statement

I have a question of the energy required to haul 2000kg 4 metres up a 1 in 100 incline with 300 Newtons friction. Velocity unknown.

Relevant Equations

Work = Force times Distance, PE = mgh

Hello. Be v grateful any assistance with this.

Question is energy to haul 2000kg 4 metres up a 1 in 100 incline with 300 Newtons friction

Answer is given as exactly 1.98 kJ.

The only figures I can see that relate to anything here are 2000kg minus 10 = 1980, and 1:100 is 0.01 which times 1000 is ten. I don't see how you put those together.

Many thanks :)

 

[berkeman]

Welcome to PhysicsForums.

Relevant Equations:: Work = Force times Distance, PE = mgh

Those are the correct relevant equations. Use the 2nd equation to tell you how much work it took to lift the mass up the vertical height of the incline. Use the 1st equation and the retarding force of friction that you are given to figure out how much extra work was needed to pull the mass up the hypotenuse of the incline. Does that make sense? Show us that work so we can check it for you.

 

[haruspex]

The only figures I can see that relate to anything here are 2000kg minus 10 = 1980, and 1:100 is 0.01 which times 1000 is ten. I don't see how you put those together.

It is 2021. Did you try using that too? (And btw, 2000-10=1990.)

It is a waste of time using numerology to solve physics problems. Even if you happen on the right answer you will have learned nothing.
As @berkeman notes, you have quoted the relevant equations, but you do not seem to know how to use them. An equation isn't really complete without definitions of the variables and a specification of the conditions under which it applies.

In Work = Force times Distance, in what direction, in relation to the force, must that distance be measured?
In PE = mgh, what is h exactly?

As so often happens, there is more than one route to the answer.
Berkeman's instruction above doesn't match those equations directly. He has separated out the total work into two parts, the work overcoming gravity and the work overcoming friction. So in applying Work = Force times Distance he would use only the frictional force, not the force opposing gravity.

An alternative is to figure out the total force to be overcome and only use Work = Force times Distance.

 

[Bix]

Thanks again for your answers

FxD = 1200 (4x300)

To find h in PE = mgh:

Either divide one by a hundred and times by 4 = 0.04

Or

Inv Tan 1/100 = 0.57 approx
Adjacent side length 4m divided by Cos 0.47 = 4.0002 = length of hypotenuse
H squared 16 and 1/625 minus A squared 16 = B squared
B = 1/25 = 0.04
B = height by which load is raised against gravity

Then

PE = 2000 x 9.8 x (B= 0.04) = 784

Ans = 1200 + 784 = 1984 kJ
But Ans = 1.98kJ

Do you round off 4 joules? Surely not..

 

[PeroK]

Do you round off 4 joules? Surely not..

What's $4J$ in the grand scheme of things?

You rounded off gravity to $9.8 m/s^2$. That's only two significant figures.

 

[PeroK]

PS That said, you may have misinterpreted 1:100 incline. That means the tangent is 0.01.

 

[Bix]

Hi PeroK thanks: the run is 4m so I'm finding a rise of 0.04

 

[PeroK]

Hi PeroK thanks: the run is 4m so I'm finding a rise of 0.04

That's not 1:100.

 

[Bix]

If it's 1:100 then when the run is 100 the rise is 1.00. So I'm reasoning that when the run is 4 the rise is 0.04.

 

[haruspex]

Adjacent side length 4m divided by Cos 0.47 = 4.0002 = length of hypotenuse

If it's 1:100 then when the run is 100 the rise is 1.00. So I'm reasoning that when the run is 4 the rise is 0.04.

The 4m is the hypotenuse. You need to find the tangent to get the rise.
But for such small angles, sine and tan are very nearly equal.

Re precision, the rule of thumb is to quote the same number of significant figures in the answer as in the input number with the fewest. That works when all the numbers are going to be combined via multiplication and division, but in this question there is also an addition, which complicates matters.
Worse, some of the precisions in the given data are unclear. Technically, "4m" could be anything from 3.5m to 4.5m. If more precision is intended then it should be given as 4.0m or 4.00m, etc.
The numbers 100, 300 and 2000 in the question, technically, also only provide one significant figure. To make those more precise requires either a ".0" on the end, or scientific notation (like 2.00x10³kg), or an explicit range...
Unfortunately, most problem setters are more cavalier and leave you to guess the intended precision.
In this problem, the one thing you can be sure is that g is only given to two figures. If we take the rest as exact then we have 1200+7.8x10²J=1.98kJ.

 

[PeroK]

The 4m is the hypotenuse. You need to find the tangent to get the rise.
But for such small angles, sine and tan are very nearly equal.

Nearly equal, but enough of a difference to account for $4J$.

 

[haruspex]

Nearly equal, but enough of a difference to account for $4J$.

I get that it makes a difference of 0.04J.

 

[Bix]

OK. Thanks very much again for your help