Questions: For an experiment at the University of Idaho, DeShuna sends an object down a tube 0.5 meters in length. The object undergoes an electromotive force given by F=5 cos (x π/2) Newtons, where x is the distance of the object from one end of the tube. How much work is required to move the object through the tube? N-m

Solution Steps

The problem requires calculating the work done to move an object through a tube under a variable force. The force is given as a function of position \( F(x) = 5 \cos \left(\frac{x \pi}{2}\right) \).

The work done by a variable force along a path is given by the integral of the force over the distance. Here, the work \( W \) is: \[ W = \int_{0}^{0.5} F(x) \, dx = \int_{0}^{0.5} 5 \cos \left(\frac{x \pi}{2}\right) \, dx \]

To solve the integral, we need to find the antiderivative of \( 5 \cos \left(\frac{x \pi}{2}\right) \). The antiderivative is: \[ \int 5 \cos \left(\frac{x \pi}{2}\right) \, dx = \frac{5}{\left(\frac{\pi}{2}\right)} \sin \left(\frac{x \pi}{2}\right) + C = \frac{10}{\pi} \sin \left(\frac{x \pi}{2}\right) + C \] Evaluate this from 0 to 0.5: \[ W = \left[ \frac{10}{\pi} \sin \left(\frac{x \pi}{2}\right) \right]_{0}^{0.5} = \frac{10}{\pi} \sin \left(\frac{0.5 \pi}{2}\right) - \frac{10}{\pi} \sin \left(\frac{0 \pi}{2}\right) \]

Calculate the values: \[ \sin \left(\frac{0.5 \pi}{2}\right) = \sin \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \] \[ \sin \left(\frac{0 \pi}{2}\right) = \sin(0) = 0 \] Substitute these into the expression for work: \[ W = \frac{10}{\pi} \left(\frac{\sqrt{2}}{2}\right) - \frac{10}{\pi} (0) = \frac{5\sqrt{2}}{\pi} \]

The work required to move the object through the tube is: \[ W = \frac{5\sqrt{2}}{\pi} \, \text{N-m} \]

Final Answer