Answer: k = ma + uk×mgcosθ/ xf
Explanation: The body is placed on a frictionless inclined ramp.
The weight of the object has 2 components, horizontal component (mgsinθ) and vertical component (mgcosθ).
The horizontal component of weight is responsible for making tje object slide down the plane even with no applied force.
So from newton's second law of motion
mgsinθ - uk×R = ma
Where uk = coefficient of kinetic friction.
R = normal reaction = mgcosθ
mgsinθ - uk×mgcosθ = ma
mgsinθ = ma + uk×mgcosθ
mgsinθ is the applied force in this case. This applied force compresses a spring.
According to hooke's law,
F =ke
Where F = ma + uk×mgcosθ, e =xf
F = applied force , e = extension and k = spring constant.
k = F/e
k = ma + uk×mgcosθ/ xf
