What is the minimum work needed to push a 1025 kg car 290 m up a 17.5° incline with zero friction?
weight force down slope = m g sin 17.5
= 1025 (9.81) sin 17.5
= 3023 N
distance = 290 in direction of needed force
so
Work = 3024 N * 290 m = 876,865 Joules
By the way
if you do it in 10 minutes which is 600 seconds
that is 876,865/600 = 1,461 watts
You need about 2 horsepower :)
To calculate the minimum work needed to push a car up an incline with zero friction, we can use the concept of work and the formula:
Work = Force × Distance × Cosine(θ)
Where:
- Force is the component of the force applied parallel to the direction of motion. It is given by Force = mass × acceleration.
- Distance is the displacement along the incline.
- θ represents the angle of the incline.
First, let's calculate the force required to move the car up the incline. The component of the force parallel to the incline is given by:
Force = mass × acceleration
Since the car is on an incline, the acceleration is given by:
Acceleration = gravitational acceleration × sine(θ)
The gravitational acceleration is 9.8 m/s^2. Therefore, the force required to move the car up the incline is:
Force = mass × gravitational acceleration × sine(θ)
Next, we need to calculate the distance along the incline. From the given information, the car is moved 290 m up the incline. Therefore, the distance is:
Distance = 290 m
Finally, we can calculate the angle of the incline. It is given as 17.5°.
Now, we have all the values needed to calculate the minimum work required. By substituting the values into the formula, we can find the answer:
Work = Force × Distance × Cosine(θ)
Work = (mass × gravitational acceleration × sine(θ)) × Distance × Cosine(θ)
Work = (1025 kg × 9.8 m/s^2 × sine(17.5°)) × 290 m × Cosine(17.5°)
Using a scientific calculator, we can evaluate this expression to find the minimum work needed to push the car up the incline.