Question

question 6 1 pts what horsepower is required to pull a 2000 pound airplane up a 20 - foot ramp onto a three - foot high platform in one minute

Explanation:

Step1: Recall Power Formula

Power \( P \) (in foot - pounds per second) is work \( W \) divided by time \( t \), \( P=\frac{W}{t} \). Work \( W \) done against gravity is \( W = F\times d\), where \( F \) is the force (weight of the airplane, 2000 lb) and \( d \) is the vertical distance (3 ft).
So \( W=2000\times3 = 6000\) foot - pounds.

Step2: Convert Time to Seconds

Time \( t = 1\) minute \(= 60\) seconds.

Step3: Calculate Power in ft - lb/s

\( P=\frac{6000}{60}=100\) foot - pounds per second.

Step4: Convert to Horsepower

We know that 1 horsepower (hp) \( = 550\) foot - pounds per second. So \( hp=\frac{100}{550}=\frac{2}{11}\approx0.18\) hp. Wait, maybe I made a mistake. Wait, the ramp is 20 - foot long, vertical height 3 ft. The force along the ramp: using similar triangles, the force \( F_{ramp}=\frac{weight\times vertical\ height}{ramp\ length}=\frac{2000\times3}{20}=300\) lb. Then work \( W = F_{ramp}\times ramp\ length=300\times20 = 6000\) foot - pounds (same as before). Time \( t = 60\) s. Power \( P=\frac{6000}{60}=100\) ft - lb/s. Then \( hp=\frac{100}{550}=\frac{2}{11}\approx0.18\) hp. Wait, maybe the question is about the power to lift, but the force along the ramp is \( F=\frac{2000\times3}{20}=300\) lb. Then power \( P = F\times v\), where velocity \( v=\frac{20\ ft}{60\ s}=\frac{1}{3}\ ft/s\). So \( P = 300\times\frac{1}{3}=100\) ft - lb/s. Convert to hp: \( \frac{100}{550}=\frac{2}{11}\approx0.18\) hp. Wait, maybe I messed up the work. Work done is against gravity, so \( W = mgh\) (in imperial units, \( F = mg = 2000\) lb, \( h = 3\) ft), so \( W=2000\times3 = 6000\) ft - lb. Time \( t = 60\) s. Power \( P=\frac{6000}{60}=100\) ft - lb/s. Since 1 hp = 550 ft - lb/s, \( hp=\frac{100}{550}=\frac{2}{11}\approx0.18\) hp.

Answer:

\(\frac{2}{11}\) (or approximately \(0.18\)) horsepower