Question

1. what is the average force given a mass of 77.0 kg? assume that at t = 0 minutes, the object is at rest. (show work on separate paper 3pts)

chart with t(min) and v(m/s): t=1 min, v=5.00 m/s; t=2 min, v=5.50 m/s; t=3 min, v=5.75 m/s; t=4 min, v=5.60 m/s; t=5 min, v=5.62 m/s
multiple-choice options: 0.204 n, 0.162 n, 1.44 n, 0.830 n

Explanation:

Step1: Recall Impulse-Momentum Theorem

The impulse-momentum theorem states that the impulse ( \( J \) ) is equal to the change in momentum ( \( \Delta p \) ). Impulse is also given by the average force ( \( F_{avg} \) ) multiplied by the time interval ( \( \Delta t \) ), and momentum is \( p = mv \). So, \( J = F_{avg}\Delta t=\Delta p = m\Delta v \). We need to find the average force, so we can rearrange to \( F_{avg}=\frac{m\Delta v}{\Delta t} \).

Step2: Determine Initial and Final Conditions

The object is at rest at \( t = 0 \) minutes, so initial velocity \( v_i = 0 \, \text{m/s} \). We need to choose a final time and velocity. Let's take the final time as \( t = 5 \) minutes. First, convert time to seconds: \( \Delta t=5\,\text{min}\times60\,\text{s/min} = 300\,\text{s} \). The final velocity at \( t = 5 \) min is \( v_f = 5.62\,\text{m/s} \). The mass \( m = 77.0\,\text{kg} \).

Step3: Calculate Change in Velocity

\( \Delta v=v_f - v_i=5.62\,\text{m/s}- 0\,\text{m/s}=5.62\,\text{m/s} \)

Step4: Calculate Average Force

Using \( F_{avg}=\frac{m\Delta v}{\Delta t} \), substitute the values: \( F_{avg}=\frac{77.0\,\text{kg}\times5.62\,\text{m/s}}{300\,\text{s}} \)

First, calculate the numerator: \( 77.0\times5.62 = 77\times5 + 77\times0.62=385+47.74 = 432.74 \)

Then divide by 300: \( F_{avg}=\frac{432.74}{300}\approx1.44\,\text{N} \) (If we take a different time, say \( t = 1 \) min: \( \Delta t = 60\,\text{s} \), \( v_f = 5.00\,\text{m/s} \), \( F_{avg}=\frac{77\times5}{60}=\frac{385}{60}\approx6.42\,\text{N} \), but that's not an option. Wait, maybe the question is about average force over the entire time? Wait, the options include 1.44 N. Let's check the calculation again. Wait, maybe I misread the time. Wait, the table has t in minutes: 1,2,3,4,5. So if we take the total time from t=0 to t=5 min, which is 5*60=300 s. Mass 77 kg, final velocity 5.62 m/s. Then \( F_{avg}=\frac{77\times5.62}{300}=\frac{432.74}{300}\approx1.44\,\text{N} \), which matches one of the options.

Answer:

\( \boxed{1.44\,\text{N}} \)