Dropped Rock Kinetic Energy Problem

AI Thread Summary
A rock weighing 40 N is dropped from a height of 20 m, and the kinetic energy just before it hits the ground can be calculated using the work-energy principle or potential energy formulas. The work done on the rock is 800 J, which equals the kinetic energy at impact, as energy is conserved without air friction. The potential energy approach yields the same result, confirming that the kinetic energy is 800 J. Calculating the velocity just before impact involves using kinematic equations, ultimately leading to the same energy value. The discussion clarifies that the rock's kinetic energy is 800 J right before it reaches the ground.
NIT14
Ok, I missed last class period, but I have the general idea of this I think. I just don't know how to put it all together for this one...

A rock that weighs 40 N is dropped from a cliff 20 m above the ground. What is the kinetic energy of the rock when it gets to the ground? (disregard air friction).

Ok, I know that F=MA, I know that KE=1/2mv^2, and I know that PE=mgh.

Am I missing something here, or am I just retarded? lol. I got the an answer of 10290J and I don't even remember how!.. and I really don't think this is correct. Can anyone help? Thanks!
 
Physics news on Phys.org
Are you considering an elastic collision. If not, the rock will not be moving after being stopped by the ground and if it is not moving then it has no KE.

Or is the question asking just before it hits the ground?

Reread the question and retype it.

Nautica
 
I'm pretty sure it means right before it gets to the ground. If not, then the energy would be converted to heat engery, right?

I don't think it has anything to do with "elastic collision" either.
 
Okay, you know the acceleration is 9.8 m/s^2. Use that to find the velocity at just before it hits the ground.

Use that velocity and plug it into you KE equation.

Does that help.

Nautica
 
Originally posted by NIT14
A rock that weighs 40 N is dropped from a cliff 20 m above the ground. What is the kinetic energy of the rock when it gets to the ground? (disregard air friction).
It's sometimes best to lead by example instead of giving hints on how to do it.

There are several ways to get the answer to this question. The easiest by far is to use the work formula.

W = Fd
W = (40N)(20m)
W = 800J

The potential energy way of looking at it is exactly the same since mg is the same as F.
E = mgh
E = (40N)(20m)
E = 800J

Then there is the kinetic energy way which is just a waste of time since you have to find the velocity and the mass. It's also less accurate because you have to do more operations.

m = F/a
m = 40/9.81
m = 4.0775kg

Vf^2 = Vi^2 + 2ad
Vf^2 = 2(9.81)(20)
Vf = 19.809m/s

E = (1/2)mv^2
E = (1/2)(4.0775)(19.809)^2
E = 799.998J


Don't pay much attention to how many significant figures I use.
 
Hey, thanks a lot guys! Makes perfect sense now. :smile:
 

Thread 'Errors and significant figures'

I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
View full post »
Thread 'Collision of a bullet on a rod-string system: query'

Thread 'Collision of a bullet on a rod-string system: query'

In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
View full post »
Thread 'A cylinder connected to a hanging mass'

Thread 'A cylinder connected to a hanging mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
View full post »

Similar threads

Finding kinetic energy and initial velocity of a cart over time
Replies
6
Views
2K
Conservation of Momentum Problem - Mechanical and Kinetic Energies
Replies
2
Views
897
Expression for magnitude of the constant friction force
Replies
7
Views
954
Minimal rotational kinetic energy for a gyroscope to precess
Replies
33
Views
2K
A block dropping onto a spring system
2
Replies
58
Views
2K
Question about the Kinetic Energy of a baseball in flight
Replies
8
Views
3K
How Fast Does a Rock Fall from 20 Meters?
Replies
1
Views
1K
How Does Dropping Rocks from Different Heights Affect Kinetic Energy Ratios?
Replies
4
Views
3K
Kinetic energy of a bouncing ball
Replies
7
Views
4K
Rotating Rod in Plane: Kinetic Energy & Moment of Inertia
Replies
16
Views
2K

Hot Threads

Recent Insights

Back
Top