Question
Samantha pushes a \( 50-\mathrm{N} \) crate up a ramp 25.0 m in length and inclined at \( 10^{\circ} \) with the horizontal. What potential energy change does the crate experience? 1) 13 J 2) 55 J 3) 120 J 4) 220 J 5) 280 J
Ask by Stanley Tucker. in the United States
Feb 25,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
The crate gains approximately 220 J of potential energy.
Solution
To calculate the change in gravitational potential energy (PE), we use the formula:
ΔPE = weight × vertical displacement
The vertical displacement can be found by multiplying the distance along the ramp by the sine of the angle of inclination:
vertical displacement = 25.0 m × sin(10°)
Using sin(10°) ≈ 0.17365,
vertical displacement ≈ 25.0 m × 0.17365 ≈ 4.341 m
Now, calculate the potential energy change:
ΔPE = 50 N × 4.341 m ≈ 217.05 J
This value is closest to 220 J, so the correct answer is:
4) 220 J
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Simplify this solution Beyond the Answer
To find the change in potential energy, we use the formula \( \Delta PE = mgh \). First, we need to calculate the height \( h \) gained by the crate as it moves up the ramp. The height can be found using the sine of the angle and the length of the ramp: \[ h = L \sin(\theta) = 25.0 \, \text{m} \times \sin(10^{\circ}) \] Calculating this gives: \[ h \approx 25.0 \, \text{m} \times 0.1736 \approx 4.34 \, \text{m} \] Now we can use the weight (force) to find the change in potential energy: \[ \Delta PE = mgh = 50 \, \text{N} \times 4.34 \, \text{m} = 217 \, \text{J} \] Looking at the options provided, the closest to our calculation is 220 J. Therefore, the answer is: 4) 220 J
