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Small objects are delivered to the 2-meter inclined chute by a conveyor belt A moving at a speed V₁ = 0.3 m/s. If the objects are delivered to belt B at V₂ = 0.9 m/s with no slipping and a coefficient of friction μₖ = 0.5 between the objects and the chute, what is the incline angle (θ) of the chute?

A) 30∘

B) 45∘

C) 60∘

D) 75∘

User Rastasheep
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Final answer:

To find the incline angle (θ) of the chute, we can use the concept of friction. The incline angle can be calculated using the equation θ = arctan(V_1 / (V_2 - μ_k * V_1)), where μ_k is the coefficient of friction and V_1 and V_2 are the speeds of conveyor belts A and B respectively.

Step-by-step explanation:

To find the incline angle (θ) of the chute, we can use the concept of friction. When an object is delivered to belt B without slipping, the friction force between the object and the chute helps to maintain its speed.

The friction force is given by the equation:

F_friction = μ_k * N

where μ_k is the coefficient of friction and N is the normal force. The normal force is equal to the weight of the object:

N = m * g

where m is the mass of the object and g is the acceleration due to gravity.

Since the object is delivered at a speed of 0.9 m/s, the friction force must be equal to the force applied by the conveyor belt (F_applied) to keep the object moving without slipping:

F_applied = F_friction

Combining these equations, we have:

μ_k * m * g = F_applied

From the given information, the coefficient of friction (μ_k) is 0.5 and the speed of conveyor belt A (V_1) is 0.3 m/s. Therefore, the incline angle (θ) can be calculated using the equation:

θ = arctan(V_1 / (V_2 - μ_k * V_1))

User Ravendra Kumar
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