PHY 151- Lab M1

VI-1 Pan with Teflon Bottom: m p = 64.2 g m (g) m t (g) m s (g) 200.0 264.2 55.0 200.0 264.2 55.0 200.0 264.2 90.0 200.0 264.2 40.0 200.0 264.2 60.0 400.0 464.2 115.0 400.0 464.2 130.0 400.0 464.2 100.0 400.0 464.2 105.0 400.0 464.2 100.0 600.0 664.2 170.0 600.0 664.2 160.0 600.0 664.2 165.0 600.0 664.2 160.0 600.0 664.2 180.0 800.0 864.2 190.0 800.0 864.2 225.0 800.0 864.2 220.0 800.0 864.2 240.0 800.0 864.2 230.0 1000.0 1064.2 260.0 1000.0 1064.2 280.0 1000.0 1064.2 260.0 1000.0 1064.2 240.0 1000.0 1064.2 290.0

Slope, S 0.2615 Uncertainty in Slope, σ s 0.01084425 Intercept, B -8.8883 Uncertainty in Intercept, σ b 7.8286318 The slope of the line can be written as S= m s m t and using the relationship m s = μ k ∙m t we can rearrange this equation to state μ k = m s m t , which would make S = μ k . Hence the slope of the graph is equal to the coefficient of kinetic friction. Additionally, the uncertainty in slope correlates to the uncertainty in μ k . Therefore, using the values from above: S = m s m t = μ k VI-2 Pan with Rubber Bottom: m p’ = 127.8 g m (g) m t (g) m s (g) 100.0 227.8 80.0 100.0 227.8 80.0 100.0 227.8 90.0 μ k ±σ μ k = 0.26 ± 0.01 g

Intercept, B 4.5364 Uncertainty in Intercept, σ b 5.7823293 The slope of the line can be written as S= m s m t and using the relationship m s = μ k ∙m t we can rearrange this equation to state μ k = m s m t , which would make S = μ k . Hence the slope of the graph is equal to the coefficient of kinetic friction. Additionally, the uncertainty in slope correlates to the uncertainty in μ k . Therefore, using the values from above: S = m s m t = μ k VI-3 Teflon Bottom: L = 59.3 cm d = 16.4 cm Mass (g) m t H (cm) Y (cm) θ ( deg. ) μ s 200.0 264.2 27.0 10.6 10.3 0.2 400.0 464.2 26.5 10.1 9.8 0.2 600.0 664.2 25.9 9.5 9.2 0.2 800.0 864.2 24.7 8.3 8.0 0.1 1000.0 1064.2 24.2 7.8 7.6 0.1 Values of Y: Values of Y were found using the equation: Y= H-d Sample Calculations: Y= H-d Y= 27.0 cm – 16.4 cm Y= 10.6 cm Values of θ : Each value of θ was determined using the values of L, H, and d. They were found using the equation θ = sin − 1 ( Y L ) . Sample Calculations: θ = sin − 1 ( Y L ) . θ = sin − 1 ( 10.6 59.3 ) μ k ±σ μ k = 0.36 ± 0.01 g

θ = 10.3 ° Values of μ s : Each value of μ s was found using the equation μ s = tan θ c . tan θ c = y x Sample Calculations: μ s = tan θ c μ s = tan ( 10.3 ) μ s = ¿ 0.2 Average Value ´ μ s : Since tan θ c is equal to μ s , the average value of μ s is the sum of all the calculated values of μ s divided by the number of masses tested which is 5. Therefore μ s = ∑ μ s 5 . Calculation: μ s = ∑ μ s 5 μ s = ∑ 0.2 + 0.2 + 0.2 + 0.1 + 0.1 5 μ s = ¿ 0.16 Standard Deviation of μ s , σ μ s : To calculate the standard deviation of μ s I used the formula s n − μ s μ ¿ ¿ ¿ n 2 ¿ ¿ ∑ ¿ ¿ σ μ s = √ ¿ . Calculation: s n − μ s μ ¿ ¿ ¿ n 2 ¿ ¿ ∑ ¿ ¿ σ μ s = √ ¿ σ μ s = √ ( 0.16 − 0.2 ) 2 +( 0.16 − 0.2 ) 2 +( 0.16 − 0.2 ) 2 +( 0.16 − 0.1 ) 2 +( 0.16 − 0.1 ) 2 5 − 1 σ μ s = 0.05

Since tan θ c is equal to μ s , the average value of μ s is the sum of all the calculated values of μ s divided by the number of masses tested which is 5. Therefore μ s = ∑ μ s 5 . Calculation: μ s = ∑ μ s 5 μ s = ∑ 0. 4 + 0.4 + 0.4 + 0.4 + 0.4 5 μ s = ¿ 0.4 Standard Deviation of μ s , σ μ s : To calculate the standard deviation of μ s I used the formula s n − μ s μ ¿ ¿ ¿ n 2 ¿ ¿ ∑ ¿ ¿ σ μ s = √ ¿ . Calculation: s n − μ s μ ¿ ¿ ¿ n 2 ¿ ¿ ∑ ¿ ¿ σ μ s = √ ¿ σ μ s = √ ( 0.4 − 0.4 ) 2 +( 0.4 − 0.4 ) 2 +( 0.4 − 0.4 ) 2 +( 0.4 − 0.4 ) 2 +( 0.4 − 0.4 ) 2 5 − 1 σ μ s = 0 .0 μ s ±σ μ s = ¿ 0.4 ± 0.0