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iren [92.7K]
3 years ago
6

A long ramp made of cast iron is sloped at a constant angle θ = 52.0∘ above the horizontal. Small blocks, each with mass 0.42 kg

but made of different materials, are released from rest at a vertical height h above the bottom of the ramp. In each case the coefficient of static friction is small enough that the blocks start to slide down the ramp as soon as they are released. You are asked to find h so that each block will have a speed of 4.00 m/s when it reaches the bottom of the ramp. You are given these coefficients of sliding (kinetic) friction for different pairs of materials.
Material 1 Material 2 Coefficient of Sliding Friction
Cast iron Cast iron 0.15
Cast iron Copper 0.29
Cast iron Lead 0.43
Cast iron Zinc 0.85
Physics
1 answer:
dezoksy [38]3 years ago
4 0

Answer:

For cast iron we have

h = 0.92 m

For copper

h = 1.05 m

For Lead

h = 1.23 m

For Zinc

h = 2.43 m

Explanation:

As we know that final speed of the block is calculated by work energy theorem

W_f + W_g = \frac{1}{2}mv^2

now we have

-\mu_k mg cos\theta(\frac{h}{sin\theta}) + mgh = \frac{1}{2}mv^2

now we have

v^2 = 2gh - 2\mu_k g h cot\theta

v = \sqrt{2gh(1 - \mu_k cot\theta)}

For cast iron we have

4 = \sqrt{2(9.81)(h)(1 - 0.15cot52)}

h = 0.92 m

For copper

4 = \sqrt{2(9.81)(h)(1 - 0.29cot52)}

h = 1.05 m

For Lead

4 = \sqrt{2(9.81)(h)(1 - 0.43cot52)}

h = 1.23 m

For Zinc

4 = \sqrt{2(9.81)(h)(1 - 0.85cot52)}

h = 2.43 m

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- a third force, given by the propulsion of the plane, which is accelerating it towards the ground (because the problem says that the plane has an acceleration of a=12 m/s^2 towards the ground)

The radius of the circle is r= \frac{260 m}{2} = 130 m, so the centripetal force acting on the plane is
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(b) The tangent of the angle with respect to the horizontal is the ratio between the sum of the forces in the vertical direction (taken with negative sign, since they are directed downward) and the forces acting in the horizontal direction, so:
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A spring gun consists of a spring inside a plastic tube with spring constant, k. The spring can be compressed 20 cm from its equ
emmainna [20.7K]

Answer: The spring constant is K=392.4N/m

Explanation:

According to hook's law the applied force F will be directly proportional to the extension e produced provided the spring is not distorted

The force F=ke

Where k=spring constant

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Ignoring air resistance

assuming g=9.81m/s²

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masha68 [24]

Answer:

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