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oksian1 [2.3K]
3 years ago
12

A car drives around a horizontal, circular track at constant speed. Consider the following three forces that act on the car: (1)

The upward normal force exerted on the car by the road, (2) the downward gravitational force on the car, (3) and the frictional force that is directed toward the center of the circular path.
Which of these forces does zero work on the car as the car moves along the circular path?  1, 2, and 3   3   1 and 2 
 1   2   1 and 3 
 2 and 3 
Physics
1 answer:
musickatia [10]3 years ago
6 0

Answer: 1, 2 and 3.

Explanation:

By definition, work, is the process through which a force, applied on an object, produces a displacement of this object, and can be expressed as the dot product of the force vector and the displacement vector.

It can be also understood as the projection of the force in the direction of the displacement, so, if both vectors are perpendicular, as the projection of one vector along the other is nul, if a force and the displacement are perpendicular, no work is done.

In our case, the 3 forces mentioned, are perpendicular respect the displacement.

For normal force and gravity, as they act vertically, and the car moves along an horizontal trajectory, both are perpendicular, so no work is done.

For the friction force, as it is the only horizontal force present, it is just the centripetal force that keeps the car moving in a circular path.

It always aims to the center of the circle, along the radius at any point, so it is always perpendicular to the displacement also.

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The horizontal force applied to the block is approximately 1,420.84 N

The known parameters;

The mass of the block, w₁ = 400 kg

The orientation of the surface on which the block rest, w₁ = Horizontal

The mass of the block placed on top of the 400 kg block, w₂ = 100 kg

The length of the string to which the block w₂ is attached, l = 6 m

The coefficient of friction between the surface, μ = 0.25

The state of the system of blocks and applied force = Equilibrium

Strategy;

Calculate the forces acting on the blocks and string

The weight of the block, W₁ = 400 kg × 9.81 m/s² = 3,924 N

The weight of the block, W₂ = 100 kg × 9.81 m/s² = 981 N

Let <em>T</em> represent the tension in the string

The upward force from the string = T × sin(θ)

sin(θ) = √(6² - 5²)/6

Therefore;

The upward force from the string = T×√(6² - 5²)/6

The frictional force = (W₂ - The upward force from the string) × μ

The frictional force, F_{f2} = (981 - T×√(6² - 5²)/6) × 0.25

The tension in the string, T = F_{f2} × cos(θ)

∴ T = (981 - T×√(6² - 5²)/6) × 0.25 × 5/6

Solving, we get;

T = \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8} \approx 183.27

Frictional \ force, F_{f2} = \left (981 -  \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8}  \times \dfrac{\sqrt{6^2 - 5^2} }{6} \times  0.25 \right) \approx 219.92

The frictional force on the block W₂, F_{f2} ≈ 219.92 N

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The force acting the block w₁, due to w₂ F_{w2} = 219.92/0.25 ≈ 879.68

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P = The horizontal force applied to the block

P = (W₁ + \mathbf{F_{w2}}) × μ + \mathbf{F_{f2}}

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