To solve the problem involving the cart on a ramp and hanging mass, we assume the system moves at constant velocity (acceleration (a=0)) and calculate the coefficient of kinetic friction using the given tension (T=4.05\,\text{N}).
For the cart ((M=1.00\,\text{kg})) on the ramp:
The net force along the ramp is zero. The forces are:
Thus:
[Mg\sin\theta = T + f\quad \text{(1)}]
The normal force (N) perpendicular to the ramp is:
[N=Mg\cos\theta\quad \text{(2)}]
Given:
(Mg\sin\theta = (1.00)(9.8)\sin30^\circ = 4.9\,\text{N})
(T=4.05\,\text{N})
From equation (1):
[f=Mg\sin\theta - T = 4.9 - 4.05 = 0.85\,\text{N}]
From equation (2):
[N=Mg\cos\theta=(1.00)(9.8)\cos30^\circ \approx 8.487\,\text{N}]
[μ_k=\frac{f}{N}=\frac{0.85}{8.487}\approx 0.100]
Answer: (\boxed{0.100}) (rounded to three significant figures)
(\boxed{0.100})
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