A block is released from rest on an inclined plane and moves 2.7m during the next 2 seconds.?

Question

I already found the magnitude of the acceleration, but I can't figure out how to calculate the coefficient of kinetic friction for the incline.
I've looked at other questions/answers, none have helped me get the right answer on this one.

Answer 1

The net force pulling and accelerating the block down along the plane is f = ma = W sin(theta) - k W cos(theta); where theta is the angle of incline and k is the coefficient you are looking for. W cos(theta) = N, the normal weight pressing into the face of the plane and W = mg, the weight of the block.

We can now write a = g sin(theta) - kg cos(theta)] from above. Therefore kg cos(theta) = g sin(theta) - a; then k = [g sin(theta) - a]/(g cos(theta)). You say you have acceleration a, g is 9.81 m/sec^2 on Earth's surface, and I presume theta is also given. Therefore you have all the values to solve for k.

Answer 2

take down the plane as the positive direction
given
d = 2.7m, t = 2s
so
a = 2d/t² = 1.35m/s²

then
the force down the plane due to gravity is mgsinθ and
the force up the plane due to friction is -μmgcosθ
where μ is the coefficient of kinetic friction
and mgcosθ is the normal force
so using newtons law
ma = ∑F
ma = mgsinθ + (-μmgcosθ)
gives
μ = (gsinθ - a) / gcosθ = calculator

,.,.,.,.,.,.,,.,.

Answer 3

By s = ut + 1/2at^2 =>2.4 = 0 + 1/2 x a x (5.3)^2 =>a = 0.17 m/s^2 =>By F(net) = F(gravity) - F(friction) =>m x a = mgsinθ - µ x N =>m x a = mgsinθ - µ x mgcosθ =>µ = [gsinθ -a]/[gcosθ] =>µ = [9.8 x sin22* - 0.17]/[9.8cos22*] =>µ = 0.39