6

Let m denote the mass of the case, friction is limiting thus: F=μR

R=mgcos(θ) hence F=μmgcos(θ)

If the block comes to a rest, the frictional force must be greater than its component of weight down the ramp. If the two forces were equal, it would continue with constant speed not come to rest: F>mgsin(θ)

μmgcos(θ)>mgsin(θ) therefore μ>tan(θ) as we can divide by cod(θ) due to 0<θ<π/2

There are various ways of doing the next part. The net force acting on the block is given by:

F(net) = mg(μcos(θ)-sin(θ)), W=Fd hence W = mgd(μcos(θ)-sin(θ))

The initial kinetic energy must have been equal to this work done so Ek=(1/2)mu^2=mgd(μcos(θ)-sin(θ))

Finally, cancelling m’s and multiplying by two then square rooting achieves:

u = [2gd(μcos(θ)-sin(θ))]^(1/2)