a motorist runs out of gas and must push his 3000kg car across a flat road 200m to the gas station. ignoring friction how much work must be done in order to get the car to the gas pump? if the motorist accomplishes this in 10 minutes, how much power has he applied? what if the coefficient of kinetic friction is u= 0.012? how much work is done? how much power is applied?
In the zero-friction case, ZERO work and no power is required on a flat (level) road.
If the coefficient of kinetic friction is 0.012, a force of
F = 0.012 M g = 353 N is required to push the car. (That is about 80 lb)
Use that for the work and power required.
Work = F * X
Power = F*X/(600 seconds)
To find the work done in pushing the car and the power applied, we need to use the formulae related to work, power, and friction. Let's break it down step by step.
Step 1: Calculate the work done ignoring friction.
When ignoring friction, the only force acting on the car is the force of pushing. The work done by a force is given by the formula:
Work = Force * Distance
In this case, the force is equal to the weight of the car, which can be calculated using the formula:
Force = Mass * Acceleration due to gravity
The distance is 200m given in the question. The mass of the car is 3000kg. The acceleration due to gravity is approximately 9.8 m/s². Plugging these values into the formula:
Force = 3000kg * 9.8 m/s² = 29400 N
Work = 29400 N * 200m = 5,880,000 J (joules)
Therefore, to push the car to the gas pump without considering friction, approximately 5,880,000 joules of work must be done.
Step 2: Calculate the power applied.
Power is defined as the amount of work done per unit of time. The formula for power is:
Power = Work / Time
The time given is 10 minutes, but it's more convenient to convert it to seconds:
10 minutes * 60 seconds/minute = 600 seconds
Plugging the values into the formula:
Power = 5,880,000 J / 600s = 9800 W (watts)
Therefore, the motorist has applied approximately 9800 watts of power.
Step 3: Calculate the work done and power applied considering friction.
When considering kinetic friction, we need to account for the additional opposing force. The formula for the work done with friction is the same:
Work = Force * Distance
However, the force now includes the force of friction, which is given by:
Force of friction = Normal force * Coefficient of kinetic friction
The normal force is the weight of the car equal to 3000kg * 9.8 m/s² = 29400 N.
The coefficient of kinetic friction given is u = 0.012.
The distance is still 200m. Plugging these values into the formula:
Force of friction = 29400 N * 0.012 = 352.8 N
Work = (29400 N + 352.8 N) * 200m = 6,078,560 J (joules)
Therefore, considering friction, approximately 6,078,560 joules of work must be done.
Step 4: Calculate the power applied with friction.
Using the same formula for power:
Power = Work / Time
Plugging in the values:
Power = 6,078,560 J / 600s = 10,130 W (watts)
Therefore, when accounting for kinetic friction, the motorist has applied approximately 10,130 watts of power.
To summarize:
- Ignoring friction, approximately 5,880,000 joules of work must be done, and approximately 9800 watts of power is applied.
- Considering friction (coefficient of kinetic friction u = 0.012), approximately 6,078,560 joules of work must be done, and approximately 10,130 watts of power is applied.