Question

A frictionless inclined plane is 6 m long and rests on a wall that is 2 m high.

Answer 1

Answer:

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

Explanation:

It is given that,

Length of the friction less plane, l = 6 m

Height of the wall, h = 2 m

Weight of the block of ice, W = 300 N

The length of the inclined plane and the height of the wall act as hypotenuse and the perpendicular of the right angled triangle. Using trigonometry to find the angle as :

$sin\theta=\dfrac{P}{H}$

$sin\theta=\dfrac{2}{6}$

$sin\theta=\dfrac{1}{3}$

Force needed to push a block of ice weighing 300 N up the plane is equal to the horizontal component of the force as :

$F_x=F\ sin\theta$

$F_x=300\times \dfrac{1}{3}$

$F_x=100\ N$

So, the force needed to push a block of ice is 100 N. Hence, this is the required solution.