a 18.5 kg penguin slides at a constant velocity of 1.40 m/s down an icy incline. The incline slopes above the horizontal at an angle of 6.90 degrees. At the bottom of the incline, the penguin slides onto a horizontal patch of ice. The coefficient of kinetic friction between the penguin and the ice is the same for the incline as for the horizontal patch. During the time the penguin slides down the slope, what is the normal force? the friction force? the coefficient of friction?
To find the normal force (N), we can start by drawing a free body diagram of the penguin on the incline:
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N | \
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mg θ
In the diagram, N is the normal force, mg is the weight of the penguin (mass * gravity), and θ is the angle of the incline.
Based on the diagram, we can see that N is the component of weight perpendicular to the incline. Since the incline slopes above the horizontal at an angle of 6.90 degrees, the component of weight perpendicular to the incline is given by N = mg * cos(θ).
Now we can calculate the normal force:
N = (18.5 kg)(9.8 m/s^2) * cos(6.90 degrees)
= (18.5)(9.8)(cos(6.90))
= 179.46 N
Therefore, the normal force acting on the penguin is approximately 179.46 N.
Next, let's find the friction force (f). The friction force between the penguin and the ice is the same for the incline as for the horizontal patch. Since the penguin is sliding down the incline at a constant velocity, the friction force is equal in magnitude and opposite in direction to the component of weight parallel to the incline.
The component of weight parallel to the incline is given by mg * sin(θ).
f = mg * sin(θ)
= (18.5 kg)(9.8 m/s^2) * sin(6.90 degrees)
= (18.5)(9.8)(sin(6.90))
= 20.46 N
Therefore, the friction force acting on the penguin is approximately 20.46 N.
Lastly, let's find the coefficient of friction (μ). The coefficient of kinetic friction (μ) relates the friction force (f) to the normal force (N) with the equation f = μN.
μ = f / N
= 20.46 N / 179.46 N
= 0.114
Therefore, the coefficient of kinetic friction between the penguin and the ice is approximately 0.114.