What is the Power Output of Gravity?

[Kyle Wies]

Homework Statement

:
It takes gravity 1.43 seconds to pull a 3.67 kg cat down from a 10m tall ledge. What is the power output of gravity? [/B]

Homework Equations

P=mgd/t
P=w/t
P=power
M=mass
G=gravity
D=distance
W=work
T=time[/B]

The Attempt at a Solution

P=(3.67)(9.81)(10)/1.43
P=251.77

 

[Ray Vickson]

Homework Statement

:
It takes gravity 1.43 seconds to pull a 3.67 kg cat down from a 10m tall ledge. What is the power output of gravity? [/B]

Homework Equations

P=mgd/t
P=w/t
P=power
M=mass
G=gravity
D=distance
W=work
T=time[/B]

The Attempt at a Solution

P=(3.67)(9.81)(10)/1.43
P=251.77

Do you have a question? Do your quantities $P, M, D, W, T$ have units? If so, you need to state them at some point.

 

[Kyle Wies]

Oops forgot to add those!
There's
N-Newtons
M-mass
J-joule
S-seconds
N*m-Newton mass

My question is what is the power output of gravity. I'm not sure if it is just 9.81 or something else. Any guidance helps!

 

[kuruman]

The power output of gravity is $P_g=\frac{dW_g}{dt}$ where $W_g$ is the work done by gravity on the object. It is not "work divided by time".

 

[Ray Vickson]

Oops forgot to add those!
There's
N-Newtons
M-mass
J-joule
S-seconds
N*m-Newton mass

My question is what is the power output of gravity. I'm not sure if it is just 9.81 or something else. Any guidance helps!

The standard value for the acceleration ($g$) of gravity (near the Earth's surface) is about 9.80665 (m/s²), which rounds to 9.81 if we want three significant figure accuracy. However, most "elementary" physics applications often take $g = 9.8 \; m/s^2$; I, personally, use 9.81.

In fact, due to "centrifugal" effects caused by the Earth's rotation, the actual value of $g$ depends on latitude:
$$ \begin{array}{lr}
g_{\rm{poles}} &=& 9.832 \; m/s^2 \\
g_{45^o} &=& 9.806 \; m/s^2\\
g_{\rm{equator}} &=& 9.780 \; m/s^2
\end{array}$$
So, unless you are doing super-accurate ballistic or aeronautical computations, you can just go ahead and use $g = 9.81$.

By the way: NEVER denote it by $G$ as you did; in Physics the symbol $G$ stands for the "gravitational constant", not the acceleration of gravity. People might very well subtract marks if you use $G$ instead of $g$.

 

[haruspex]

The power output of gravity is $P_g=\frac{dW_g}{dt}$ where $W_g$ is the work done by gravity on the object. It is not "work divided by time".

@Kyle Wies , in case you are confused by that post, let me clarify.
The question ought to have asked what the average power output was over the given time. You have correctly answered that.
@kuruman's point is that the power is not constant. Since the force is constant, it increases in proportion to the velocity.

 

[haruspex]

due to "centrifugal" effects caused by the Earth's rotation

To be accurate, the centrifugal effect changes apparent gravity, not the actual force exerted. The reduction at the equator is about 0.3%. In addition, because of the shape of the Earth, the radius is greater at the equator, reducing the actual force too - by about 0.2% compared with the poles. So the total reduction in apparent gravity is about 0.5%.