Simple force/energy/work problem

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In summary, the minimum work needed to push a 950kg car 310m up along a 9.0 degree incline is 9.81 m/s2.
  • #1
xecure
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simple force & work problem

I'm not very good as physics and the question i was having problems with is this: what is the minimum work needed to push a 950kg car 310m up along a 9.0 degree incline? (ignore friction)


the formula i was using is this: Work = Force * distance * cos(theta)

distance and theta was already given so i used F=ma to try and find the force needed, but the acceleration for the object is not given, and neither is velocity (so i am not able to use: [PLAIN]https://www.physicsforums.com/latex_images/90/905663-3.png)


What formula should i use? how should i go about doing this?
 
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  • #2
You do the minimum work when pushing the car with a force that just balances the downward force of gravity along the slope, and do not waste work for acceleration. Well, you need to accelerate the car at the beginning so as it starts to move, but then just keep this low speed.

ehild
 
  • #3
ehild said:
You do the minimum work when pushing the car with a force that just balances the downward force of gravity along the slope, and do not waste work for acceleration. Well, you need to accelerate the car at the beginning so as it starts to move, but then just keep this low speed.

ehild

I'm stumped, if acceleration does not come into play how would i go about doing this problem?
 
  • #4
How much force have you exert to prevent a car rolling down a slope? You need just a bit more to push it up. Your work is this force multiplied by the length of the slope.
 
  • #5
ehild said:
How much force have you exert to prevent a car rolling down a slope? You need just a bit more to push it up. Your work is this force multiplied by the length of the slope.

but I am having trouble finding the force, how would i go about calculating the force?
 
  • #6
If you hold a bag of 10 kg what is the force you exert on the bag?

If you were so strong that you could lift that car of 950 kg and hold it, how much would be your force?

ehild
 
  • #7
it would be m*a, which is 950kg * 9.81 = 9319.5
 
  • #8
xecure said:
it would be m*a, which is 950kg * 9.81 = 9319.5

Does that car accelerate if it is in rest?

9.81 m/s2 is the acceleration of a body when it falls. Is the car falling? Acceleration means change of velocity. If a body is in rest, the velocity is zero, the acceleration is zero. If the body moves with constant velocity, the acceleration is zero.

The gravity of Earth is a force that acts on every body. It causes downward acceleration g on the free bodies. If you exert force equal to gravity but opposite to it, these forces will cancel and the car stays in rest or moves with constant velocity.

Gravity does work on a falling body. If the body falls from height h, the work of gravity is Wg=mgh. When you lift a body with zero acceleration, your force is equal to gravity, F=mg. Lifting the car at height h needs W=mgh work from you.

The slope is 310 m long and the angle of the incline is 9°. What is the height then?
Of course, you do not lift the car just vertically up, but it is pushed uphill. If there is no friction, no loss of energy, the work is the same either the car is moved up on the slope or raised up vertically : W=mgh.

ehild
 

FAQ: Simple force/energy/work problem

1. What is a simple force/energy/work problem?

A simple force/energy/work problem is a type of physics problem that involves calculating the work done by a force on an object. It typically involves determining the amount of force applied, the distance the object moves, and the resulting work done.

2. How do you calculate work in a simple force/energy/work problem?

Work is calculated by multiplying the force applied to an object by the distance the object moves in the direction of the force. The formula for work is W = F * d, where W is work, F is force, and d is distance.

3. What units are used to measure work in a simple force/energy/work problem?

The SI unit for work is the joule (J). However, other units such as foot-pounds (ft-lb) and ergs (erg) may also be used.

4. Can you have negative work in a simple force/energy/work problem?

Yes, negative work can occur if the force is applied in the opposite direction to the motion of the object. This means that the work done is in the opposite direction of the displacement, resulting in a negative value.

5. How does energy relate to work in a simple force/energy/work problem?

Work and energy are closely related in a simple force/energy/work problem. Work is the transfer of energy from one object to another, and the amount of work done is equal to the change in energy. In other words, the work done on an object is equal to the energy gained by that object.

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