Moving up an incline requires extra energy to counter the force of gravity.
It reduces the amount of lifting force and also reduces the wind energy of all of the substances and molecules in the prototype (the item that you're moving)
Lengthen the ramp, decrease the mass of the object, use a machine (e.g., a block and tackle) to pull the weight up the ramp, reduce the friction of the weight against the ramp, move the ramp further from the center of gravity of the earth, submerge the ramp in a liquid...tbere may be more ways but this should give you some ideas to consider.
More force will be required to push an object along the ramp.
The cart stays where it is and doesn't move.
A
It reduces the amount of lifting force and also reduces the wind energy of all of the substances and molecules in the prototype (the item that you're moving)
An Egyptian ramp is a ramp with a platform in the middle to reduce the incline of the ramp and/or to change the direction of incline.
The applied force will depend on the required force, and the angle to the ramp (or the horizontal) at which the force is applied.
A ramp exerts no force, just gravity.
Lengthen the ramp, decrease the mass of the object, use a machine (e.g., a block and tackle) to pull the weight up the ramp, reduce the friction of the weight against the ramp, move the ramp further from the center of gravity of the earth, submerge the ramp in a liquid...tbere may be more ways but this should give you some ideas to consider.
No. It reduces the amount of force required, but it does not reduce the total amount of work. In fact, due to friction, it will probably increase the total amount of work.
More force will be required to push an object along the ramp.
Assuming 100% efficiency, the amount of work depends on the weight and the vertical distance (that is, opposite to the pull of gravity) moved. If you use a ramp then you reduce the force by an amount k, that's true, but the distance you have to push in the direction of the force is multiplied by k. The work done is [original force]/k times [vertical distance]times k and the k cancels out. For vertical lifting k=1.
The cart stays where it is and doesn't move.
Reduce the friction of it and the ramp, for example, mounting it on wheels.
The input force would increase as the height of the ramp increased. It wouldn't matter the distance. Ask me another one.
The gravitational force is.