Question

A frictionless inclined plane is 8.0 m long and rests on a wall that is 2.0 m high. How much force is needed to push a block of ice weighing 300.0 N up the plane

Answer 1

Given Information:  

Inclined plane length = 8 m

Inclined plane height = 2 m

Weight of ice block = 300 N

Required Information:  

Force required to push ice block = F = ?  

Answer:  

Force required to push ice block =  75 N

Explanation:  

The force required to push this block of ice on a inclined plane is given by

F = Wsinθ

Where W is the weight of the ice block and θ is the angle as shown in the attached image.

Recall from trigonometry ratios,

sinθ = opposite/hypotenuse

Where opposite is height of the inclined plane and hypotenuse in the length of the inclined plane.

sinθ = 2/8

θ = sin⁻¹(2/8)

θ = 14.48°

F = 300*sin(14.48)

F = 75 N

Therefore, a force of 75 N is required to push this ice block on the given inclined plane.