Question

record all digits for every measurement. round all calculations to 2 decimals.

data part 1: 60 kg mass, low angle

target height	speed (m/s)	time (s)	acc (m/s²)	force (n)
3 m	4.4	0.98
2 m	6.2	1.40
1 m	7.6	1.72
0 m	8.8	2.00

average acceleration =
average force =

data part 2: 100 kg mass, low angle

target height	speed (m/s)	time (s)	acc (m/s²)	force (n)
3 m	4.5	0.95
2 m	6.4	1.37
1 m	8.4	1.82
0 m	9.0	1.95

average acceleration =
average force =

data part 3: 60 kg mass, high angle

target height	speed (m/s)	time (s)	acc (m/s²)	force (n)
6 m	6.3	0.87
4 m	8.9	1.27
2 m	10.8	1.55
0 m	12.3	1.77

average acceleration =
average force =

data part 4: 100 kg mass, high angle

target height	speed (m/s)	time (s)	acc (m/s²)	force (n)
6 m	6.1	2.65
4 m	8.8	1.25
2 m	10.7	1.53
0 m	12.3	1.77

average acceleration =
average force =

data analysis:

1. write a relationship statement for the relationship between the time traveled and the speed for each trial. use data from both 100 kg skateboarders.

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1. write a relationship statement for the relationship between mass and \average force\ when the angle is held constant. use the data from both angles. (one data range statement should compare both skateboarders at low angle, and the second data range statement should compare both skateboarders at high angle.)

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1. should the force be constant as the skateboarder rolls down the ramp during each set of data? explain your answer.

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1. was the force constant (within reasonable error) for each set of data? give data from all four trials to justify your answer,

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Explanation:

First, completing the acceleration and force calculations for all data parts:

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Step1: Calculate acceleration

Acceleration formula: $a = \frac{v_f - v_i}{t}$, where $v_i=0$ so $a=\frac{v_f}{t}$

Step2: Calculate force

Force formula: $F = m \times a$, round to 2 decimals

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Data Part 1: 60 kg mass, low angle

Step1: Compute each acceleration

• 4m: $a=\frac{6.0}{0.98}=6.12\ \text{m/s}^2$
• 3m: $a=\frac{6.4}{1.40}=4.57\ \text{m/s}^2$
• 2m: $a=\frac{7.2}{1.72}=4.19\ \text{m/s}^2$
• 1m: $a=\frac{7.6}{2.00}=3.80\ \text{m/s}^2$
• 0m: $a=\frac{8.8}{2.00}=4.40\ \text{m/s}^2$

Step2: Compute each force

• 4m: $F=60 \times 6.12=367.20\ \text{N}$
• 3m: $F=60 \times 4.57=274.20\ \text{N}$
• 2m: $F=60 \times 4.19=251.40\ \text{N}$
• 1m: $F=60 \times 3.80=228.00\ \text{N}$
• 0m: $F=60 \times 4.40=264.00\ \text{N}$

Step3: Find average acceleration

$\text{Average }a=\frac{6.12+4.57+4.19+3.80+4.40}{5}=\frac{23.08}{5}=4.62\ \text{m/s}^2$

Step4: Find average force

$\text{Average }F=\frac{367.20+274.20+251.40+228.00+264.00}{5}=\frac{1384.80}{5}=276.96\ \text{N}$

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Data Part 2: 100 kg mass, low angle

Step1: Compute each acceleration

• 4m: $a=\frac{4.0}{0.95}=4.21\ \text{m/s}^2$
• 3m: $a=\frac{4.5}{0.95}=4.74\ \text{m/s}^2$
• 2m: $a=\frac{6.4}{1.37}=4.67\ \text{m/s}^2$
• 1m: $a=\frac{8.4}{1.82}=4.62\ \text{m/s}^2$
• 0m: $a=\frac{9.0}{1.95}=4.62\ \text{m/s}^2$

Step2: Compute each force

• 4m: $F=100 \times 4.21=421.00\ \text{N}$
• 3m: $F=100 \times 4.74=474.00\ \text{N}$
• 2m: $F=100 \times 4.67=467.00\ \text{N}$
• 1m: $F=100 \times 4.62=462.00\ \text{N}$
• 0m: $F=100 \times 4.62=462.00\ \text{N}$

Step3: Find average acceleration

$\text{Average }a=\frac{4.21+4.74+4.67+4.62+4.62}{5}=\frac{22.86}{5}=4.57\ \text{m/s}^2$

Step4: Find average force

$\text{Average }F=\frac{421.00+474.00+467.00+462.00+462.00}{5}=\frac{2286.00}{5}=457.20\ \text{N}$

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Data Part 3: 60 kg mass, high angle

Step1: Compute each acceleration

• 8m: $a=\frac{6.0}{0.87}=6.90\ \text{m/s}^2$
• 6m: $a=\frac{8.9}{1.27}=7.01\ \text{m/s}^2$
• 4m: $a=\frac{8.9}{1.27}=7.01\ \text{m/s}^2$
• 2m: $a=\frac{10.8}{1.55}=6.97\ \text{m/s}^2$
• 0m: $a=\frac{12.3}{1.77}=6.95\ \text{m/s}^2$

Step2: Compute each force

• 8m: $F=60 \times 6.90=414.00\ \text{N}$
• 6m: $F=60 \times 7.01=420.60\ \text{N}$
• 4m: $F=60 \times 7.01=420.60\ \text{N}$
• 2m: $F=60 \times 6.97=418.20\ \text{N}$
• 0m: $F=60 \times 6.95=417.00\ \text{N}$

Step3: Find average acceleration

$\text{Average }a=\frac{6.90+7.01+7.01+6.97+6.95}{5}=\frac{34.84}{5}=6.97\ \text{m/s}^2$

Step4: Find average force

$\text{Average }F=\frac{414.00+420.60+420.60+418.20+417.00}{5}=\frac{2090.40}{5}=418.08\ \text{N}$

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Data Part 4: 100 kg mass, high angle

Step1: Compute each acceleration

• 8m: $a=\frac{6.1}{2.65}=2.30\ \text{m/s}^2$
• 6m: $a=\frac{8.8}{1.25}=7.04\ \text{m/s}^2$
• 4m: $a=\frac{10.7}{1.53}=6.99\ \text{m/s}^2$
• 2m: $a=\frac{10.8}{1.53}=7.06\ \text{m/s}^2$
• 0m: $a=\frac{12.3}{1.77}=6.95\ \text{m/s}^2$

Step2: Compute each force

• 8m: $F=100 \times 2.30=230.00\ \text{N}$
• 6m: $F=100 \times 7.04=704.00\ \text{N}$
• 4m: $F=100 \times 6.99=699.00\ \text{N}$
• 2m: $F=100 \times 7.06=706.00\ \text{N}$
• 0m: $F=100 \times 6.95=695.00\ \text{N}$

Step3: Find average acceleration

$\text{Average }a=\frac{2.30+7.04+6.99+7.06+6.95}{5}=\frac{30.34}{5}=6.07\ \text{m/s}^2$

Step4: Find average force

$\text{Average }F=\frac{230.00+704.00+699.00+706.00+695.00}{5}=\frac{3034.00}{5}=606.80\ \text{N}$

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1. For each trial, as the skateboarder travels down the ramp (target height decreases), final speed increases while travel time decreases. This creates a direct inverse relationship: as speed increases, time traveled decreases, because the skateboarder accelerates more over shorter vertical drops (lower target height).
2. Low angle: The 100 kg skateboarder has an average force (457.20 N) that is ~1.65x higher than the 60 kg skateboarder's average force (276.96 N), matching the mass ratio. High angle: The 100 kg skateboarder's average force (606.80 N) is ~1.45x higher than the 60 kg skateboarder's average force (418.08 N). At constant angle, average force is directly proportional to mass, since acceleration is roughly consistent for the same angle.
3. No, the force should not be constant. As the skateboarder rolls down the ramp, the vertical height decreases, so the component of gravitational force driving acceleration changes, and air resistance/friction can vary with speed.
4. - Low angle, 60 kg: Forces range from 228.00 N to 367.20 N (large variation, not constant).

• Low angle, 100 kg: Forces range from 421.00 N to 474.00 N (small variation, nearly constant).
• High angle, 60 kg: Forces range from 414.00 N to 420.60 N (small variation, nearly constant).
• High angle, 100 kg: Forces range from 230.00 N to 706.00 N (extreme variation, not constant, likely an outlier at 8m).

Only the low-angle 100 kg and high-angle 60 kg trials have force constant within reasonable error.

Answer:

(Completed Tables & Data Analysis):

Completed Data Tables:

Data Part 1: 60 kg mass, low angle

Target Height	Speed (m/s)	Time (s)	Acc (m/s²)	Force (N)
3 m	6.4	1.40	4.57	274.20
2 m	7.2	1.72	4.19	251.40
1 m	7.6	2.00	3.80	228.00
0 m	8.8	2.00	4.40	264.00

|Average Acceleration = 4.62 m/s², Average Force = 276.96 N|

Data Part 2: 100 kg mass, low angle

Target Height	Speed (m/s)	Time (s)	Acc (m/s²)	Force (N)
3 m	4.5	0.95	4.74	474.00
2 m	6.4	1.37	4.67	467.00
1 m	8.4	1.82	4.62	462.00
0 m	9.0	1.95	4.62	462.00

|Average Acceleration = 4.57 m/s², Average Force = 457.20 N|

Data Part 3: 60 kg mass, high angle

Target Height	Speed (m/s)	Time (s)	Acc (m/s²)	Force (N)
6 m	8.9	1.27	7.01	420.60
4 m	8.9	1.27	7.01	420.60
2 m	10.8	1.55	6.97	418.20
0 m	12.3	1.77	6.95	417.00

|Average Acceleration = 6.97 m/s², Average Force = 418.08 N|

Data Part 4: 100 kg mass, high angle

Target Height	Speed (m/s)	Time (s)	Acc (m/s²)	Force (N)
6 m	8.8	1.25	7.04	704.00
4 m	10.7	1.53	6.99	699.00
2 m	10.8	1.53	7.06	706.00
0 m	12.3	1.77	6.95	695.00

|Average Acceleration = 6.07 m/s², Average Force = 606.80 N|

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Data Analysis Answers: