A block of mass 2 kg slides on an inclined plane which makes an angle 30 degree with the horizontal coefficient of friction between the block and the surface is √(3/2) what food should be applied to the block so that the block moves down without any acceleration
Answers
Answered by
22
Hey dear,
◆ Answer-
F = 20.77 N
◆ Explanation-
# Given-
m = 2 kg
μ = √(3/2) = 1.225
θ = 30 °
# 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.
Hope it's helpful...
◆ Answer-
F = 20.77 N
◆ Explanation-
# Given-
m = 2 kg
μ = √(3/2) = 1.225
θ = 30 °
# 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.
Hope it's helpful...
Similar questions