A can of sardines is made to move along an x axis from x1 = 0.2 m to x2 = 1.2 m by a force given by F = iF0exp(-0.4x), with x in meters and F0 = 4.5 N. (Here exp is the exponential function.) How much work is done on the can by the force?
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* Physics/Math - drwls, Thursday, March 1, 2007 at 2:43am
Work is the integral of force x distance.
In this case,
------1.2
W = (Integral)4.5 exp(-0.4x)dx
------0.2
= 4.5/(-0.4)[exp(-0.4*1.2) - exp(-0.4*0.2)]
>>>>>>>>>>>>>>>>>>>>>>>>>>
Ok, so according to this W = 36.97 J
This, however, is the wrong answer. Did I do something wrong in my calculations???
Significant figures: the least precise x is in ONE significant figure. Geez. The most precise is two sif figures. You have the problem set up correctly.
I don't get the same final value as you, we are different by approximately a factor of ten.
I got it. 3.42. thanks
how do you get F0= 4.5N? My problem only says:
...by a force with a magnitude given by
F = exp(–3x), with x in meters and F in newtons...
am i missing any info by the program or i have to find that out?
To calculate the work done on the can by the force, we need to integrate the force over the distance traveled. Here's how you can do it step by step:
1. Write down the expression for the force: F = iF0exp(-0.4x), where i is the unit vector along the x-axis, F0 = 4.5 N, and x is the position along the x-axis in meters.
2. Set up the integral for the work done: W = ∫F dx, where the integral is taken from x1 = 0.2 m to x2 = 1.2 m.
3. Substitute the expression for the force into the integral: W = ∫iF0exp(-0.4x) dx.
4. Evaluate the integral: W = F0∫exp(-0.4x) dx.
5. Integrate the exponential function: W = F0 * (-1/0.4) * exp(-0.4x) + C, where C is the constant of integration.
6. Evaluate the integral between the limits of integration: W = F0 * (-1/0.4) * [exp(-0.4x2) - exp(-0.4x1)].
7. Plug in the values for x1 and x2: W = F0 * (-1/0.4) * [exp(-0.4*1.2) - exp(-0.4*0.2)].
8. Evaluate the exponential terms: W = 4.5 * (-1/0.4) * [exp(-0.48) - exp(-0.08)].
9. Simplify the expression: W = 11.25 * [0.617383 - 0.923116].
10. Calculate the difference: W = 11.25 * (-0.305733).
11. Finally, calculate the work done: W = -3.440%).
So the correct value for the work done on the can by the force is approximately -3.44 J (Joules). It looks like you made a calculation error in step 9 or 10, which led to the wrong answer.
To find the work done on the can by the force, you need to evaluate the integral of the force function over the distance the can moves.
Let's calculate step by step:
1. Determine the work done formula:
W = ∫(F dx)
2. Substitute the force function:
W = ∫(iF0 exp(-0.4x) dx)
3. Integrate with respect to x:
W = iF0/(-0.4) ∫(exp(-0.4x) dx)
4. Integrate using the integral of the exponential function:
W = iF0/(-0.4) [-1/0.4 exp(-0.4x)]
5. Evaluate the integral limits at x = 1.2 and x = 0.2:
W = iF0/(-0.4) [-1/0.4 exp(-0.4*1.2) - (-1/0.4 exp(-0.4*0.2))]
6. Simplify and calculate:
W = (4.5 N)/(-0.4) [-2.5 exp(-0.48) + 2.5 exp(-0.08)]
Using a calculator or software, calculate the exponential values:
exp(-0.48) ≈ 0.6197
exp(-0.08) ≈ 0.9241
Now substitute these values into the equation and compute the result:
W ≈ (4.5 N)/(-0.4) [-2.5*0.6197 + 2.5*0.9241]
W ≈ (11.25/0.4)*[-1.5493 + 2.3102]
W ≈ 28.125*(0.7609)
W ≈ 21.431 J
So, the correct value for work done on the can by the force is approximately 21.431 J.