If the normal force exerted by the track on the car when it is at the top of the track (point BB) is 6.00 NN , what is the normal force on the car when it is at the bottom of the track (point AA)?

Explanation:

We already know that the wight in reference is gravitational force on an object:.

Gravity=

Weight of car=0.800kg

To solve for A, we apply Newton's second law of radial direction:

Rewrite to calculate acceleration,:

Solving for Position A:

r=rho l/a where a is the area, l is the length and rho is resistivity so the greater the area the lesser the resistance thats why the earth pin is thicker bcos current follow the shortest path, so less resistive .

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards