Question

Need help with work problem: Question attached. Please explain.

Answer 1

There are two components to this problem. The work done during a constant force for the first 6.88m and then the work done during the linearly decreasing force for the second 6.88m. The first part is easy: \[\huge \text{W}=\vec{F} \cdot \vec{d}=|\vec{F}||\vec{d}| \cos(0)\]Second part we must graph in order to formulate a modeled equation for the force, and then we must integrate since force is the derivative of work.

You already have the graph, so I'm not going to draw it again unless you need me to. But if you look, you can calculate the slope as: \[m=\frac{\Delta y}{\Delta x}= -\frac{60}{6.88}\]We also see that our y-intercept is 60N, so plugging into the simple algebraic equation for a line (in slope intercept) we get:\[f(x)=mx+b \implies F(x)=- \frac{ 60 }{ 6.88 }x+60\] Now that we have a function for our force over the second interval of distance. Time to integrate\[\text{W}=\int\limits_0^{6.88} - \frac{ 60 }{ 6.88 }x+60 ~dx\]Technically, our limits are supposed to be from 6.88 to 13.76, but since it's linear, we can make it easier and start at 0 so there is less algebra. Now we have to put the two together to account for both intervals of distance
\[\huge \text{W}=|\vec{F}||\vec{d}|+ \int\limits F(x)~dx\]I will leave the rest to you

Answer 2

Let me know if you have more questions