part 1: fan carts\nobjective: determine how the mass of an object affects its acceleration when subject to a…

part 1: fan carts\nobjective: determine how the mass of an object affects its acceleration when subject to a constant force.\nusing the lab software, we will be able to see a graph of the carts velocity over time. how would you calculate the acceleration of the cart as it is pushed by the fan, using the velocity - over - time?\nmake a prediction: increasing the mass of the cart should (increase / decrease / not change) its acceleration.\nknowing the acceleration and mass of the cart, what equation can we use to calculate the force acting on it?\ndata and analysis\n| | cart mass (kg) | change in velocity (m/s) | time (s) | calculated acceleration (m/s²) | calculated force (n) |\n|----|----|----|----|----|----|\n| trial 1 | 0.54 kg | 0.19 m/s | 1.5 s | 0.127 m/s² | 0.069 n |\n| trial 2 | 1.00 kg | 0.16 m/s | 3.1 s | 0.052 m/s² | 0.052 n |\n| trial 3 | 1.50 kg | 0.10 m/s | 4.5 s | 0.022 m/s² | 0.033 n |\nuse your data or calculations to answer the central question: how does increasing the mass of an object affect its acceleration?\ndraw a free - body diagram for the cart, showing the weight, normal force, and fan force (f_f).\nusing your data, show your calculations for the average force from the fan.
Answer
Explanation:
Step1: Calculate acceleration
Acceleration $a=\frac{\Delta v}{\Delta t}$, where $\Delta v$ is change in velocity and $\Delta t$ is time. For Trial - 1: $a_1=\frac{0.18}{1.56}\ m/s^{2}\approx0.115\ m/s^{2}$. For Trial - 2: $a_2 = \frac{0.14}{3.10}\ m/s^{2}\approx0.045\ m/s^{2}$. For Trial - 3: $a_3=\frac{0.10}{4.52}\ m/s^{2}\approx0.022\ m/s^{2}$.
Step2: Calculate force
According to Newton's second - law $F = ma$, where $m$ is mass and $a$ is acceleration. For Trial - 1: $F_1=0.54\times0.115\ N\approx0.062\ N$. For Trial - 2: $F_2 = 1.00\times0.045\ N=0.045\ N$. For Trial - 3: $F_3=1.50\times0.022\ N = 0.033\ N$.
Step3: Calculate average force
The average force $\bar{F}=\frac{F_1 + F_2+F_3}{3}=\frac{0.062 + 0.045+0.033}{3}\ N=\frac{0.14}{3}\ N\approx0.047\ N$
Answer:
The average force from the fan is approximately $0.047\ N$