The gravitational force is.
In school mechanics it does not. The force acting on the car is directly proportional to its mass (its weight adjusted for the incline of the ramp). The acceleration of the car is inversely proportional to this force. The overall result is that the mass of the car does not affect its motion. In more advanced mechanics, where friction and drag are taken into account, things start getting more complicated.
Yes, it affects the friction.
18 feet
The length of the downward sloping side of the ramp would be 18 feet approximately this is pythagarus theorem of the sum of squares. the answer should in fact be the number corresponding to the square root of (10*10+15*15).
It should be like 25 feet long and put at a like 45 degrees angle
Gravity
A ramp exerts no force, just gravity.
The heavy the crayon, it will roll down but it needs force to move it back up
The cart stays where it is and doesn't move.
friction
Even though gravity wants to pull the boy and the load strait down the wedge shape under him and the load causes a higher amount of resistance by creating a downward force while going up. This force is a help to the boy and load going down.
It has 3
The weight of the car will cause the force of gravity to push it down with seemingly more force than the lighter car. And so this would make the car go faster down a slanted surface, which is the ramp.
gravity pulls it down....but of course inertia
The applied force will depend on the required force, and the angle to the ramp (or the horizontal) at which the force is applied.
Resolve the weight into x and y componentsThe x component is the force which pushes it down and causes it to acclerateThe y component gives the normal forceFn=mgcos(A)Then the force acting sidewards Fx=mgsin(A)Total force = mgcos(A)-frictional forceFrictional force = umgsin(A)Therefore total force = mgcos(A)-umgsin(A)=mam[gcos(A)-ugsin(A)]=maso a=gcos(A)-ugsin(A)gcos(A)-a=ugsin(A)u=(gcos(A)-a)/gsin(A)
More force will be required to push an object along the ramp.