Question

A block of mass 2kg slides on an inclined plane that makes angle 30 with horizontal the coeffcient of friction between the block and surface is √3÷2 what is the force should be applied to the block so that moves down without any accelration

Answer 1

$\rule{200}{2}$

$\huge\boxed{\boxed{\underline{\mathcal{\red{A}\green{N}\pink{S}\orange{W}\blue{E}\pink{R:-}}}}}$

$\bf\large\underline\red{Explanation:-}$

# Given-

m = 2 kg

μ = √(3/2) = 1.225

θ = 30 °

$\bf\large\underline\red{Solution:-}$

For acceleration to be zero, force should be provided to balance frictional force.

F = μmgcosθ

F = 1.225 × 2 × 9.8 × cos30°

F = 20.77 N

Force of 20.77 N should be applied to get zero acceleration.

Answer 2

Answer:

Explanation:−

# Given-

m = 2 kg

μ = √(3/2) = 1.225

θ = 30 °

\bf\large\underline\red{Solution:-}

Solution:−

For acceleration to be zero, force should be provided to balance frictional force.

F = μmgcosθ

F = 1.225 × 2 × 9.8 × cos30°

F = 20.77 N

Force of 20.77 N should be applied to get zero acceleration.