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This is a photograph of an apple and feather free fall in a evacuated chamber. The apple and feather are released from the top. Suppose the camera opened flash every 1 s.

Suppose A1 is the distance from t = 0s to t = 1 s, A2 is the distance from t = 1s to t =2s, A3 is the distance from t = 2s to t = 3s ……

what is the ratio of A1: A2: A3…….

(2) If camera opened flash every 0.06s, is the ratio of the distance between every 0.06 s the same as above or change? Why?

Answer: The ratio is 1:3:5 :7……(2n -1), here n = 1, 2, 3,

- Show it in general case, suppose the time interval is Δt
- The distance from 0 –one Δt, A1 = ½*g*(Δt)2
- (notice here I consider the distance, so I neglect (-) sign)
- The distance from Δt---2 Δt, A2 = ½*g*(2Δt)2 -½*g*(Δt)2
- = ½*g*(Δt)2 (22 -1)
- Using the same way the distance from 2Δt---3 Δt,
- A3 = ½*g*(Δt)2 (32 - 22)
- A4 = ½*g*(Δt)2 (42 - 32) …….
- An = ½*g*(Δt)2 (n2 – (n-1)2)
- = ½*g*(Δt)2 (n2 – n2 +2n-1)
- = ½*g*(Δt)2 (2n-1)
- Therefore the ratio should be 1:3:5…….(2n-1), the ratio is independent of Δt

quiz2 3,

(1) A simple method of measuring the static coefficients of friction: Suppose a block is placed on a rough surface inclined relative to the horizontal as show in the Fig. The incline angle is increased until the block stars

to move. Measuring the critical angle θc at which this slipping just occurs, you can obtain μsProof: μs=tan θc

(2) The 10.2 Kg block is held in place by massless rope passing over two massless, frictionless pulleys. Find the tensions T1, T2, T3, T4, T5 and magnitude of force F.

Solution: 3,(1) If the block is not moving, Newton second law applied to the block. For this balanced situation

When the incline angle is increased until the block

Is on the verge of slipping, the force of static

Friction has reached its maximum value

The angle in this situation is the critical angle θc, in that situation, fsmax = μs N= μs mgcosθc = mg sinθc

Solve the above equations, μs N= μs mgcosθc = mg sinθc

- μs=tan θc

(2) 3,

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