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## New answers tagged integration

3

You may be imagining that if you push with constant force $F$, the spring will compress until the spring has such a resistive force. But since the spring was not counteracting that force, your constant force $F$ was accelerating the mass. Upon reaching the point where the spring has force $F$ as well, the mass does not stop but has a speed such that $KE = ... 0 The factor$\frac{1}{2}$is due to the integral. The wrong sign of yours is due to the fact that you have to counter the force of the spring. So the Force if the Spring is$-kx$, but you have to pull in the direction it is extended, so apply the force$kx$, therefore the energy is positive$W=\frac 1 2 k^2 x$5 Since the force is a function of distance, you need to integrate: $$F = kx\\ W = \int F\ dx\\ W = \int k\ x\ dx\\ W = \frac12kx^2$$ Add signs as needed... Your work considered the force to be constant - and that's not how springs work. 3 The accelerometer measures the negative of gravity plus any upwards acceleration see NOTE#1 $$acc = -(g+\ddot{x})$$ and you want to integrate$\ddot{x}$to get speed$v=\dot{x}$and position$x$. So your expressions should be $$v(t)=-\int_0^t ( acc+g)\,{\rm d}t \\ x(t)=-\int_0^t \int_0^t ( acc+g)\,{\rm d}t\,{\rm d}t$$ You also know that the final ... 2 Like Ross said, you need to try to remove the overall constant part of your acceleration data. Most accelerators report gravity all of the time. So even if your phone was sitting still on a desk, it would report and acceleration of -9.8 m/s^2 "down", however down lined up with the axes of your phone. Looking at your data a bit, for the first 1/3 of your ... 2 If your accelerations are in$g$, you should multiply the second term in column C by$9.8$. The velocities will then be in m/sec and your integration to make D will be in meters. You may then want to apply any constraints you know of. If you know the velocity starts and ends at zero, you might want to add or subtract a constant acceleration to the data to ... 3 Hints to the sought-for formula (16) for$\hat{H}$: Use integration by parts in${\bf r}$-space to remove derivatives from the Dirac delta distributions, cf. comment by user ACuriousMind. Work on the problem from both ends (15) and (16). Use Leibniz rule $$\tag{*}\nabla^2 (fg)~=~ g\nabla^2 f + f \nabla^2 g+ 2 \nabla f\cdot\nabla g,$$ so that$\nabla\$ only ...

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