Calculating the Force Required to Push a Block of Ice up an Inclined Plane

How much force is needed to push a block of ice weighing 300 N up a 6 m long frictionless inclined plane that rests on a 2 m high wall?

A) 600 N
B) 200 N
C) 300 N
D) 50 N
E) 100 N

Answer:

The force needed to push a block of ice is 100 N.

It is given that the length of the frictionless plane is 6 m and it rests on a wall that is 2 m high. The block of ice weighs 300 N. To calculate the force needed to push the block of ice up the inclined plane, we can use trigonometry.

The length of the inclined plane and the height of the wall form a right-angled triangle. By using the given values, we can find the angle of the incline and then determine the horizontal component of the force needed.

First, we find the angle theta using the sine function:
sin(theta) = height / hypotenuse
sin(theta) = 2 / 6
sin(theta) = 1 / 3

The force needed to push the block of ice weighing 300 N up the plane is equal to the horizontal component of the force:
F_x = F * sin(theta)
F_x = 300 * (1/3)
F_x = 100 N

Therefore, the force needed to push a block of ice weighing 300 N up the frictionless inclined plane is 100 N.