What is the minimum work needed to push a 900 kg car 460 m up along a 17.5° incline?
(a) Ignore friction.
(b) Assume the effective coefficient of friction retarding the car is 0.30?
Greg or Sara or any of your other many aliases -- I've removed the other 15 or so physics questions because we don't accept "homework dumping."
You're welcome to repost some questions with your own answers. Someone then, may check them.
The minimum work required in the potential energy increase at the top of the ramp. you need the height of the ramp for that. That would be 460 sin 17.5
With friction, you need to add the work done against friction, which is (friction force) x 460 m. Use the coefficient of friction and the usual formula (which you should know or learn) for the friction force
wow, my apology. im sorry.
i was expecting to get explainations and get the right equations. im very interesting in physics and i found this great website. you all are very helpful!
so i thought i could get as much help as i could.
AND you said that i can repost some questions with my answers? well i don't know how to find the answer and i thought you would show me how to do it and give me the right equations for me to use.
still want me to repost????
lol this is kinda hard. i go to glenbard north which has one of the best physics programs for high school students and we never covered this.
To find the minimum work needed to push the car up an incline, we can use the formula:
Work = Force × Distance × cos(θ),
where:
- Work is the amount of work done (in joules, J),
- Force is the force applied to push the car up the incline (in newtons, N),
- Distance is the distance moved up the incline (in meters, m),
- θ (theta) is the angle of the incline (in degrees).
(a) Without friction (Ignore friction):
Since we are ignoring friction, the only force acting on the car is the force needed to overcome the gravitational force. The gravitational force can be calculated as:
Force_gravity = mass × gravity,
where:
- mass is the mass of the car (in kilograms, kg),
- gravity is the acceleration due to gravity (approximately 9.8 m/s²).
In this case, the angle of the incline (θ) is 17.5°. To find the force needed to push the car up the incline, we can use the component of the gravitational force parallel to the incline:
Force = Force_gravity × sin(θ).
Substituting the values into the equation, we get:
Force = (900 kg × 9.8 m/s²) × sin(17.5°).
Once we have the force required, we can use it to calculate the work done:
Work = Force × Distance × cos(θ).
Substituting the given values, we get:
Work = (Force) × (460 m) × cos(17.5°).
(b) With friction (assume coefficient of friction = 0.30):
In this case, we need to consider the force due to friction. The force due to friction can be calculated as:
Force_friction = friction coefficient × Force_normal,
where:
- friction coefficient is the coefficient of friction (0.30 in this case),
- Force_normal is the normal force acting on the car.
The normal force is the force exerted by the surface perpendicular to the incline. It can be calculated as:
Force_normal = mass × gravity × cos(θ).
Once we have the force due to friction, we subtract it from the force required to overcome gravity:
Force = (Force_gravity × sin(θ)) - Force_friction.
Then, we can calculate the work done as before:
Work = Force × Distance × cos(θ).
Substituting the given values, we get:
Work = (Force) × (460 m) × cos(17.5°).
Please note that the final answer will differ depending on whether friction is considered or ignored.