Answer:
180 fb*lb
45 ft*lb
Step-by-step explanation:
We have that the work is equal to:
W = F * d
but when the force is constant and in this case, it is changing.
therefore it would be:
[tex]W = \int\limits^b_ a {F(x)} \, dx[/tex]
Where a = 0 and b = 30.
F (x) = 0.4 * x
Therefore, we replace and we would be left with:
[tex]W = \int\limits^b_a {0.4*x} \, dx[/tex]
We integrate and we have:
W = 0.4 / 2 * x ^ 2
W = 0.2 * (x ^ 2) from 0 to 30, we replace:
W = 0.2 * (30 ^ 2) - 0.2 * (0 ^ 2)
W = 180 ft * lb
Now in the second part it is the same, but the integral would be from 0 to 15.
we replace:
W = 0.2 * (15 ^ 2) - 0.2 * (0 ^ 2)
W = 45 ft * lb
Following are the calculation to the given value:
Given:
[tex]length= 30 \ ft\\\\mass= 0.4 \ \frac{lb}{ft}\\\\edge= 80 \ ft \\\\[/tex]
To find:
work=?
Solution:
Using formula:
[tex]\to W=fd[/tex]
[tex]\to W=\int^{30}_{0} 0.4 \ x\ dx\\\\[/tex]
[tex]= [0.4 \ \frac{x^2}{2}]^{30}_{0} \\\\= [\frac{4}{10} \times \frac{x^2}{2}]^{30}_{0} \\\\= [\frac{2}{10} \times x^2]^{30}_{0} \\\\= [\frac{1}{5} \times x^2]^{30}_{0} \\\\= [\frac{x^2}{5}]^{30}_{0} \\\\= [\frac{30^2}{5}- 0] \\\\= [\frac{900}{5}] \\\\=180[/tex]
Therefore, the final answer is "[tex]180\ \frac{ lb}{ft}[/tex]".
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