Physics equations, help?

This is probably really simple but I'm baffled!
The question is: 'What is the increase in Ep when a mass of 40kg is lifted 8m vertically?'
So I done 40kg x 10n/kg x 8m = 3200j
But the book says the answer is 320j, Why? I'm really confused

Reply 1

rayquaza17

Original post by courtneyr_98

Ignore my previous answer that I have now deleted (I've just failed at physics).

GPE = mgh
GPE = 40 * 10 * 8 = 3200

(edited 10 years ago)

Reply 2

rayquaza17

Original post by courtneyr_98

But the equation in the book is Ep = m x g x h
Doesn't 'g' stand for gravitational field, this on earth is 10?
I think I'm over thinking things haha, thanks

Sorry, I've changed my answer.

Yes g is 10 on the surface of the Earth.

And it is mgh.

I think your book is wrong.

Reply 3

courtneyr_98

If my revision guide is wrong what chance do I have? haha, thanks :smile:

Reply 4

VioletPhillippo

Original post by courtneyr_98

If my revision guide is wrong what chance do I have? haha, thanks :smile:

Yeah I think your revision guide is wrong too, the answer is definitely 3200j :smile:

Reply 5

l1lvink

Yeah, GPE=mgh, with g being 9.81 or 10. So i think your book just has a typo

Reply 6

samthemiller

what unit does your book give it in?

maybe it gives it in centijoules or something stupid?

Reply 7

skunkboy

Typo.

Posted from TSR Mobile

Reply 8

lerjj

Original post by samthemiller

what unit does your book give it in?

maybe it gives it in centijoules or something stupid?

Shouldn't that be hectojoules? It would be 10 x 3200J= 32 000cJ otherwise... The book has a typo probably, no reason to distrust it.

Reply 9

courtneyr_98

Yeah the book defiantly gives it in joules, aww well thanks, atleast I know I done it right :smile:

Reply 10

lawrencefrape1

Unless this action is taking place on a planet with a gravitational field strength of 1, as a pose to 10N/kg on the earth, the book is wrong. Otherwise are you sure it doesn't say 4kg, 0.8m or 3200J in the book?

Reply 11

Grecko

Here is what I have just been reading from my textbook on free falling objects:

"It is well known that, in the absence of air resistance, all objects dropped near the
Earth’s surface fall toward the Earth with the same constant acceleration under
the influence of the Earth’s gravity. It was not until about 1600 that this conclusion
was accepted. Before that time, the teachings of the Greek philosopher Aristotle
(384–322 BC) had held that heavier objects fall faster than lighter ones.
The Italian Galileo Galilei (1564–1642) originated our present-day ideas concerning
falling objects. There is a legend that he demonstrated the behavior of falling
objects by observing that two different weights dropped simultaneously from
the Leaning Tower of Pisa hit the ground at approximately the same time. Although
there is some doubt that he carried out this particular experiment, it is well established
that Galileo performed many experiments on objects moving on inclined
planes. In his experiments, he rolled balls down a slight incline and measured the
distances they covered in successive time intervals. The purpose of the incline was
to reduce the acceleration, which made it possible for him to make accurate measurements
of the time intervals. By gradually increasing the slope of the incline,
he was finally able to draw conclusions about freely falling objects because a freely
falling ball is equivalent to a ball moving down a vertical incline.
You might want to try the following experiment. Simultaneously drop a coin and
a crumpled-up piece of paper from the same height. If the effects of air resistance
are negligible, both will have the same motion and will hit the floor at the same
time. In the idealized case, in which air resistance is absent, such motion is referred to as free-fall motion. If this same experiment could be conducted in a vacuum, in
which air resistance is truly negligible, the paper and the coin would fall with the
same acceleration even when the paper is not crumpled. On August 2, 1971, astronaut
David Scott conducted such a demonstration on the Moon. He simultaneously
released a hammer and a feather, and the two objects fell together to the lunar surface.
This simple demonstration surely would have pleased Galileo!
When we use the expression freely falling object, we do not necessarily refer to an
object dropped from rest. A freely falling object is any object moving freely under
the influence of gravity alone, regardless of its initial motion. Objects thrown
upward or downward and those released from rest are all falling freely once they
are released. Any freely falling object experiences an acceleration directed downward,
regardless of its initial motion.
We shall denote the magnitude of the free-fall acceleration, also called the acceleration
due to gravity, by the symbol g. The value of g decreases with increasing altitude
above the Earth’s surface. Furthermore, slight variations in g occur with changes
in latitude. At the Earth’s surface, the value of g is approximately 9.80 m/s2. Unless
stated otherwise, we shall use this value for g when performing calculations. For
making quick estimates, use g 5 10 m/s2.
If we neglect air resistance and assume the free-fall acceleration does not vary
with altitude over short vertical distances, the motion of a freely falling object moving
vertically is equivalent to the motion of a particle under constant acceleration in
one dimension. Therefore, the equations developed in Section 2.6 for the particle
under constant acceleration model can be applied. The only modification for freely
falling objects that we need to make in these equations is to note that the motion
is in the vertical direction (the y direction) rather than in the horizontal direction
(x) and that the acceleration is downward and has a magnitude of 9.80 m/s2.
Therefore, we choose ay 5 2g 5 29.80 m/s2, where the negative sign means that
the acceleration of a freely falling object is downward."