Adisa pulls a \( 40-\mathrm{N} \) crate up a \( 5.0-\mathrm{m} \) long inclined plane at a constant velocity. If the plane is inclined at an angle of \( 37^{\circ} \) to the horizontal and there is a constant force of friction of 10 N between the crate and the surface, what is the net change in potential energy of the crate? 1) 120 J 2) -120 J 3) 200 J 4) -200 J 5) 140 J

To find the net change in potential energy of the crate, we need to calculate the height it is raised on the inclined plane. The height \( h \) can be determined using the sine function based on the angle of inclination: \[ h = d \cdot \sin(\theta) = 5.0 \, \text{m} \cdot \sin(37^{\circ}) \] Using the sine value of \( 37^{\circ} \approx 0.6 \): \[ h \approx 5.0 \, \text{m} \cdot 0.6 \approx 3.0 \, \text{m} \] Now, we can calculate the increase in potential energy \( \Delta PE \): \[ \Delta PE = mgh \] Since weight \( W = mg = 40 \, \text{N} \): \[ \Delta PE = 40 \, \text{N} \cdot 3.0 \, \text{m} = 120 \, \text{J} \] Thus, the net change in potential energy of the crate is **120 J** (Answer 1). This means that Adisa has upped the ante with that crate by raising its height—and energy—by 120 joules on that inclined plane. Impressive, right? Now, let’s talk about where this principle plays out in the real world. Whether you're moving heavy furniture into a newly decorated room or lugging gear up a camping expedition, understanding inclined planes helps us calculate how much energy we expend. Just picture the strain of lifting that hefty crate directly upwards versus sliding it up an incline! Harnessing the power of angles not only saves effort but adds a scientific twist to your everyday challenges. Who knew physics could make life a little easier—and more fun—one incline at a time?