Net Work Done by an Accelerating Brick

Homework Statement

A 12 kg brick moves along an x axis. Its acceleration as a function of its position is shown in Fig. 7-37. What is the net work performed on the brick by the force causing the acceleration as the brick moves from x = 0 to x = 5.2 m?
http://edugen.wiley.com/edugen/courses/crs4957/art/qb/qu/c07/fig07_37.gif

Homework Equations

W=integral from xo to x Fxdx + integral yo to y Fydy....

The Attempt at a Solution

Since acceleration isn't constant we can't use W=Fd, so:
Statement of known values:
xo= 0 m
x=5.2 m
m=12 kg
a(5.2)=13 ms-2 (from the graph.. determined by 4m/10ms-2=5.2m/x)
-----------------------
Unknown/want to find:
Wnet

So using the integral formula (can we assume it's only moving in the x direction?)

W=[integral]F(x)Δx

I'm at a loss though, in the example done in class, he broke the forces up in to components, but I can't really do that here since acceleration isn't constant and force is what i'm solving for...

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gneill
Mentor
"A 12 kg brick moves along an x axis". So it's one dimensional motion and you won't need components.

You may not be able to use w = f*d, but you can always integrate dw = f dot dx.

The acceleration graph appears to be a straight line, so you should be able to write an equation for the force as a function of x without too much difficulty.

"A 12 kg brick moves along an x axis". So it's one dimensional motion and you won't need components.

You may not be able to use w = f*d, but you can always integrate dw = f dot dx.

The acceleration graph appears to be a straight line, so you should be able to write an equation for the force as a function of x without too much difficulty.
I'm rather uncertain on how to use the dot product formula correctly for work, but here's a shot:

From the graph, a(x)=2x/5
v(x)=x2/5
s(x)=x3/15

so then the formula to find work done is
W=integral xo to x of x3/15 dx ?

So W= x4/60 - xo4/60
W= (5.2)4/60 - (0)4/60
W=12.19 Joules?

gneill
Mentor
You'll want to check the formula you got for the acceleration versus x. If you plug in x=4 you should get a = 10 m/s2.

With the acceleration a(x), you can write the expression for the force by using Newton's second law (f = m*a). That force is what you'll integrate from x = 0 to x = 5.2m.

$$W = \int_0^{5.2m} f(x) \cdot dx$$

You'll want to check the formula you got for the acceleration versus x. If you plug in x=4 you should get a = 10 m/s2.

With the acceleration a(x), you can write the expression for the force by using Newton's second law (f = m*a). That force is what you'll integrate from x = 0 to x = 5.2m.

$$W = \int_0^{5.2m} f(x) \cdot dx$$

I don't think I follow why you chose x=4meters, but if you do that and use Newton's second law,

F=(12kg)(10ms-2)=120N

So $$W = \int_0^{5.2m} 120 \cdot dx$$

W= 120x-120xo= 120(5.2)-120(0)= 624 Joules ?

cepheid
Staff Emeritus
Gold Member
No no no. The point is to get acceleration as a FUNCTION of x: a(x). Then, from that you can get force as a function of x: F(x). Then you can integrate the function F(x) to get the total work done. The force is NOT constant.

gneill was just saying that you should plug in a test value (for EXAMPLE x = 4) to check and see if your a(x) is correct.

No no no. The point is to get acceleration as a FUNCTION of x: a(x). Then, from that you can get force as a function of x: F(x). Then you can integrate the function F(x) to get the total work done. The force is NOT constant.

gneill was just saying that you should plug in a test value (for EXAMPLE x = 4) to check and see if your a(x) is correct.
Ohhh, so I would use the equation of the line from the graph of acceleration vs displacement

a(x)=(5/2)x
(to test) a(x)=(5/2)(4)=10

F(x)=ma(x)=(12kg)(2.5)(x)= 30x

$$W = \int_0^{5.2m} f(x) \cdot dx$$ =$$W = \int_0^{5.2m} 30x \cdot dx$$

W=15x2-15xo2 =15(5.2)2-30(0)=405.6 Joules

Last edited:
cepheid
Staff Emeritus
Gold Member
Ohhh, so I would use the equation of the line from the graph of acceleration vs displacement

a(x)=(5/2)x
(to test) a(x)=(5/2)(4)=10

F(x)=ma(x)=(12kg)(2.5)(x)= 30x

$$W = \int_0^{5.2m} f(x) \cdot dx$$ =$$W = \int_0^{5.2m} 30x \cdot dx$$

W=15x2-15xo2 =15(5.2)2-30(0)=405.6 Joules
Yeah, that looks much better.

Yeah, that looks much better.
Amazing!!! Thank you both very much! The dot product for work makes a ton and a half more sense now