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 :
[tex]sin\theta=\dfrac{P}{H}[/tex]
[tex]sin\theta=\dfrac{2}{6}[/tex]
[tex]sin\theta=\dfrac{1}{3}[/tex]
Force needed to push a block of ice weighing 300 N up the plane is equal to the horizontal component of the force as :
[tex]F_x=F\ sin\theta[/tex]
[tex]F_x=300\times \dfrac{1}{3}[/tex]
[tex]F_x=100\ N[/tex]
So, the force needed to push a block of ice is 100 N. Hence, this is the required solution.
