### Purpose

To determine the relationship between force, mass, and acceleration using a cart attached to a pulley with varying weights.

### Hypothesis

If the mass of the weights attached to the pulley is increased, the force exerted on the cart and the acceleration of the cart will also increase.

### Labeled Diagrams

See attached sheet.

### Data

Weight of Cart and Sensor (N) |
6.624 |
|||||

Weight of Cart System (kg) |
0.6759 |
|||||

Weight of Pulley (g) |
60 |
50 |
40 |
30 |
20 |
10 |

Force Exerted on Cart (N) |
0.5472 |
0.4921 |
0.4059 |
0.3227 |
0.2321 |
0.1404 |

Average Acceleration of Cart (m/s^2) |
0.8252 |
0.736 |
0.7222 |
0.4469 |
0.3395 |
0.2148 |

Weight of Cart and Sensor and 300 g (N) |
9.718 |
|||||

Weight of Cart System (kg) |
0.9916 |
|||||

Weight of Pulley (g) |
60 |
50 |
40 |
30 |
20 |
10 |

Force Exerted on Cart (N) |
0.588 |
0.5049 |
0.418 |
0.3323 |
0.2385 |
0.1442 |

Average Acceleration (m/s^2) |
0.6655 |
0.5182 |
0.4263 |
0.3478 |
0.2633 |
0.1476 |

### Graphs

See attached sheets.

### Questions

**1. Is the graph of force vs. acceleration for the cart a straight line? If so, what is the value of the slope?**

Yes, the graph produces nearly a straight line; the correlation for a linear fit is 0.9787. The value of the slope is 0.6129 N/(m/s^2).

**2. What are the units of the slope of force vs. acceleration graph? Simplify the units of the slope to fundamental units (m, kg, s). What does the slope represent?**

The units of the slope of force vs. acceleration are N/(m/s^2). This simplifies to kg. The slope represents the mass of the pulley.

**3. What is the total mass of the system (both with and without extra weight) that you measured?**

The total mass of the system without the extra weight was 0.6759 kg and the mass of the system with the extra weight (300 g) was 0.9916 kg, which seems to make sense. 0.9916 kg is almost exactly 300 g more than 0.6759 kg.

**4. How does the slope of your graph compare (percent difference) with the total mass of the system that you measured?**

The slope of the graph without any added weights was 0.6129 kg, which is a 9.78% difference. The slope of the graph with the added weights was 0.8939 kg, which is a 10.36% difference.

**5. Are the net force on an object and the acceleration of the object directly proportional?**

Yes, as the net force is increased, the acceleration is also increased.

**6. Write a general equation that relates all three variables: force, mass, and acceleration.**

F = ma

### Conclusion

**Lab Summarized**

The overall goal of the lab was to determine and show the relationship between force, mass, and acceleration. The goal was achieved using a cart and pulley system with varying weights to measure force and acceleration. The forces and accelerations collected were then graphed against each other the construct a linear fit line, whose slope showed the mass of the system (the cart, sensor, and any added weights). This value could then be compared to the mass calculated from the force of the free hanging system. The force and acceleration from each trial run could also be analyzed to show any relationship between the two values.

The data collected seemed to show a direct correlation between force and acceleration. Thus, the stated hypothesis was confirmed that if the force was increased, the acceleration would also increase. The compared values for the masses of the cart systems were about 10% different in each case. For the trial without any added weight, the calculated value of 0.6759 kg is 9.78% different from the extrapolated value of 0.6129 kg. In regards to the trial with the added weight, the calculated value of 0.9916 kg is 10.36% different than the extrapolated value of 0.8939 kg. This error could have been caused by a number of factors. For instance, the air resistance from the weight on the pulley dropping could have caused error, and any possible friction from the track could have attributed to this, too. It could also be thought that if the pulley did not drop straight downward, i.e. it was swaying at all, this would have further error. Lastly, if the rope was not completely taught when the system was put in motion, that could have caused error as well.

### Equations

F = ma

a = 9.8 m/s^2

N/(m/s^2) = kg