Question

frq practice #2: a spider sits on web with a motion detector directly beneath the spider. the spider begins to descend from rest on a single thread of web while the motion detector collects data about the motion of the spider as it descends downward. the graph is shown below. note: use $g = 10.0\\ m/s^2$. note: positive velocity on the graph below is the downward direction 4. draw a free body diagram for the spider below: 5. the mass of the spider is 0.020 kg. calculate the following based on this information and the graph. a. the weight of the spider. (hint: $f_g = mg$) answer: b. the acceleration of the spider as it moves downward. answer: c. the net force acting on the spider as it moves downward. answer: d. the tension force acting on the spider as it moves downward. answer:

Explanation:

Part 5a:

Step1: Identify formula and values

We use the formula \( F_g = mg \), where \( m = 0.020 \, \text{kg} \) and \( g = 10.0 \, \text{m/s}^2 \).

Step2: Substitute and calculate

Substitute the values into the formula: \( F_g = 0.020 \, \text{kg} \times 10.0 \, \text{m/s}^2 \).
Calculating gives \( F_g = 0.20 \, \text{N} \).

Step1: Recall acceleration from v-t graph

Acceleration is the slope of the velocity - time graph. The formula for slope is \( a=\frac{\Delta v}{\Delta t} \).
From the graph, at \( t = 0 \), \( v = 0 \); at \( t = 5 \, \text{s} \), \( v = 20 \, \text{m/s} \).

Step2: Calculate acceleration

\( a=\frac{20 \, \text{m/s}- 0 \, \text{m/s}}{5 \, \text{s}-0 \, \text{s}}=\frac{20}{5}=4.0 \, \text{m/s}^2 \)

Step1: Recall net force formula

The net force formula is \( F_{net}=ma \), where \( m = 0.020 \, \text{kg} \) and \( a = 4.0 \, \text{m/s}^2 \) (from part 5b).

Step2: Substitute and calculate

\( F_{net}=0.020 \, \text{kg}\times4.0 \, \text{m/s}^2 = 0.080 \, \text{N} \)

Answer:

\( 0.20 \, \text{N} \)

Part 5b: