A wagon was used to move a 90 lb rock 120 yds. It took 40 sec and 70 lbs of force. Can someone help me solve for efficiency and horsepower????? PLEASE :)
Power expended (in ft/lb/s)
= Force (lb) x Velocity (ft/s)
= 70 lb x (360 ft/40 s) = 630 ft/s
Convert this to horsepower using
550 ft-lb/s per HP. It's a bit more than 1 HP
The efficiency is zero unless the rock is being moved uphill, because all work is being done to overcome friction. The potential energy of the rock (and the wagon) remain the same
The ramp is 3 feet high
Sure, I can help you solve for efficiency and horsepower. Let's start with efficiency first.
Efficiency is the ratio of the useful work output to the total work input. In this case, the useful work output is moving the 90 lb rock over a distance of 120 yds, and the total work input is the force (in pounds) exerted multiplied by the distance (in yards) covered.
To calculate the total work input, you multiply the force by the distance:
Total work input = Force × Distance
Total work input = 70 lbs × 120 yds
Next, we need to calculate the useful work output. Since the wagon moved the rock, we can assume that the useful work output is equal to the weight of the rock multiplied by the distance covered:
Useful work output = Weight × Distance
Useful work output = 90 lbs × 120 yds
Now that we have both the total work input and the useful work output, we can calculate the efficiency using the equation:
Efficiency = (Useful work output / Total work input) × 100
Substituting the values:
Efficiency = (90 lbs × 120 yds) / (70 lbs x 120 yds) × 100
By simplifying this equation, we find that the distance, weight, and units (yds and lbs) will cancel out, leaving us with a unitless percentage.
Now let's move on to calculating horsepower.
Horsepower is a unit of power, which represents the rate at which work is done or energy is transferred. In this case, we need to calculate the power exerted by the wagon to move the rock.
Power is calculated using the equation:
Power = (Work / Time), where work is the total work input and time is the time taken to do the work.
Substituting the values:
Power = (70 lbs x 120 yds) / 40 sec
Now, this will give you the power in terms of pound-yards per second. If you want to convert it into horsepower, you can use the conversion factor that 1 horsepower is equivalent to 550 foot-pounds per second.
So, to convert power from pound-yards per second to horsepower:
Horsepower = (Power / 550 foot-pounds per second) × 1
I hope this explanation helps you solve for efficiency and horsepower. Let me know if you have any further questions!