a father pulls hi children in a wagon for 25 minutes covering 1.25 miles on a level sidewalk. he pulls with an average force of 35 lbf at an angle of 45 degrees to the sidewalk.
(a) how much work is done (in ft.lbf)?
(b) what is the average horsepower required?
work= force*distance*cosine45=
work= 35*.707*1.25*5280
horspower: work in ft-lbs/550
To find the work done by the father pulling the wagon, we can use the formula:
Work = Force * Distance * cos(theta)
where:
Force = 35 lbf (force applied by the father)
Distance = 1.25 miles
theta = 45 degrees (angle with the sidewalk)
(a) Calculating the work done:
First, we need to convert the given values to consistent units. Since the distance is given in miles and we need to convert it to feet, we know that 1 mile is equal to 5280 feet.
Distance = 1.25 miles * 5280 feet/mile = 6600 feet
Now, let's calculate the work done:
Work = 35 lbf * 6600 feet * cos(45 degrees)
Using a calculator:
Work = 35 lbf * 6600 feet * 0.707 (cos(45 degrees) = 0.707)
Work = 16,061.1 ft.lbf
Therefore, the work done by the father pulling the wagon is approximately 16,061.1 ft.lbf.
(b) To find the average horsepower required, we can use the formula:
Horsepower = (Work / Time) / 550
where:
Work = 16,061.1 ft.lbf (calculated in part (a))
Time = 25 minutes
First, we need to convert the time from minutes to seconds:
Time = 25 minutes * 60 seconds/minute = 1500 seconds
Now, let's calculate the average horsepower required:
Horsepower = (16,061.1 ft.lbf / 1500 seconds) / 550
Using a calculator:
Horsepower = 16,061.1 ft.lbf / 825,000 / 550
Horsepower = 0.0234 horsepower
Therefore, the average horsepower required to pull the wagon is approximately 0.0234 horsepower.
To find the answer to both parts of the question, we need to use the formulas for work and power, and make appropriate conversions.
(a) Work is defined as the product of force and displacement. The formula for work is:
Work = Force x Displacement x cos(theta)
where:
- Force is the pulling force (35 lbf)
- Displacement is the distance covered (1.25 miles)
- theta is the angle between the force and the displacement (45 degrees)
To calculate the work, we need to convert the given values to the appropriate units. 1 mile is equal to 5280 feet, and 1 lbf (pound-force) is equal to 32.174 ft.lbf (foot-pound force).
Displacement = 1.25 miles x 5280 feet/mile = 6600 feet
Force = 35 lbf x 32.174 ft.lbf/lbf = 1126.815 ft.lbf
Now, we can calculate the work:
Work = 1126.815 ft.lbf x 6600 feet x cos(45 degrees)
To find the value of cos(45 degrees), we know that cos(45 degrees) = sqrt(2)/2, which is approximately 0.7071.
Work = 1126.815 ft.lbf x 6600 feet x 0.7071 = 5246880 ft.lbf
Therefore, the work done is approximately 5246880 ft.lbf.
(b) Power is the rate at which work is done and is defined as:
Power = Work / Time
We know the work done is 5246880 ft.lbf, and the time is given as 25 minutes.
To calculate the average horsepower required, we need to make a unit conversion. 1 horsepower (hp) is equal to 550 ft.lbf/s.
Power = 5246880 ft.lbf / (25 minutes x 60 seconds/minute)
First, convert minutes to seconds:
Power = 5246880 ft.lbf / (1500 seconds)
Then, convert the result to horsepower:
Power = 5246880 ft.lbf / (1500 seconds) x (1 hp / 550 ft.lbf/s)
Power ≈ 8.023 hp
Therefore, the average horsepower required is approximately 8.023 hp.