5. a rock with a mass of 2.3 kg is dropped from a height of ten meters. the rock falls for 1.43 seconds before it hits the ground.
a. what is the acceleration of the rock as it falls?
b. what force causes this acceleration?
c. determine the rock’s velocity as it hits the ground.
d. what is the rock’s momentum as it hits ground?
e. how much kinetic energy does the rock have as it hits the ground?
f. how much potential energy does the rock have before it is dropped?
(there is a velocity-time graph on the right side of the questions with velocity (m/s) on the y-axis and time (s) on the x-axis, showing a linear plot from (0,0) to around (1.4, -14) approximately)

Question

5. a rock with a mass of 2.3 kg is dropped from a height of ten meters. the rock falls for 1.43 seconds before it hits the ground.
   a. what is the acceleration of the rock as it falls?
   b. what force causes this acceleration?
   c. determine the rock’s velocity as it hits the ground.
   d. what is the rock’s momentum as it hits ground?
   e. how much kinetic energy does the rock have as it hits the ground?
   f. how much potential energy does the rock have before it is dropped?
   (there is a velocity-time graph on the right side of the questions with velocity (m/s) on the y-axis and time (s) on the x-axis, showing a linear plot from (0,0) to around (1.4, -14) approximately)

Accepted Answer

### Part a
# Explanation:
## Step1: Identify the kinematic equation
The rock is dropped, so initial velocity $ u = 0 \, \text{m/s} $. The displacement $ h = 10 \, \text{m} $, time $ t = 1.43 \, \text{s} $. The kinematic equation for displacement with constant acceleration is $ h=ut + \frac{1}{2}at^{2} $. Since $ u = 0 $, the equation simplifies to $ h=\frac{1}{2}at^{2} $.
## Step2: Solve for acceleration $ a $
Rearrange the equation $ h=\frac{1}{2}at^{2} $ to solve for $ a $: $ a=\frac{2h}{t^{2}} $. Substitute $ h = 10 \, \text{m} $ and $ t = 1.43 \, \text{s} $ into the formula: $ a=\frac{2\times10}{(1.43)^{2}}=\frac{20}{2.0449}\approx9.78 \, \text{m/s}^2 $.

# Answer: $ \approx 9.78 \, \text{m/s}^2 $

### Part b
# Brief Explanations:
When an object falls near the Earth's surface, the acceleration is due to the gravitational pull of the Earth. The force causing this acceleration is the gravitational force (or weight of the object).
# Answer: Gravitational force (or Weight)

### Part c
# Explanation:
## Step1: Choose the kinematic equation
We can use $ v = u+at $ (since we know $ u = 0 $, $ a\approx9.78 \, \text{m/s}^2 $, $ t = 1.43 \, \text{s} $) or $ v^{2}=u^{2}+2ah $ ( $ u = 0 $, $ a\approx9.78 \, \text{m/s}^2 $, $ h = 10 \, \text{m} $ ). Let's use $ v^{2}=u^{2}+2ah $.
## Step2: Calculate the velocity
Substitute $ u = 0 $, $ a\approx9.78 \, \text{m/s}^2 $, $ h = 10 \, \text{m} $ into $ v^{2}=2ah $. $ v=\sqrt{2\times9.78\times10}=\sqrt{195.6}\approx14 \, \text{m/s} $ (or using $ v = at=9.78\times1.43\approx14 \, \text{m/s} $).

# Answer: $ \approx 14 \, \text{m/s} $ (downward)

### Part d
# Explanation:
## Step1: Recall the formula for momentum
Momentum $ p=mv $, where $ m = 2.3 \, \text{kg} $ (mass of the rock) and $ v\approx14 \, \text{m/s} $ (velocity from part c).
## Step2: Calculate the momentum
Substitute $ m = 2.3 \, \text{kg} $ and $ v = 14 \, \text{m/s} $ into $ p = mv $: $ p=2.3\times14 = 32.2 \, \text{kg·m/s} $ (downward).

# Answer: $ 32.2 \, \text{kg·m/s} $ (downward)

### Part e
# Explanation:
## Step1: Recall the formula for kinetic energy
Kinetic energy $ KE=\frac{1}{2}mv^{2} $, where $ m = 2.3 \, \text{kg} $ and $ v\approx14 \, \text{m/s} $.
## Step2: Calculate the kinetic energy
Substitute $ m = 2.3 \, \text{kg} $ and $ v = 14 \, \text{m/s} $ into $ KE=\frac{1}{2}mv^{2} $: $ KE=\frac{1}{2}\times2.3\times(14)^{2}=\frac{1}{2}\times2.3\times196 = 2.3\times98=225.4 \, \text{J} $.

# Answer: $ 225.4 \, \text{J} $

### Part f
# Explanation:
## Step1: Recall the formula for gravitational potential energy
Gravitational potential energy $ PE = mgh $, where $ m = 2.3 \, \text{kg} $, $ g\approx9.78 \, \text{m/s}^2 $ (or $ 9.8 \, \text{m/s}^2 $), $ h = 10 \, \text{m} $.
## Step2: Calculate the potential energy
Substitute $ m = 2.3 \, \text{kg} $, $ g = 9.8 \, \text{m/s}^2 $, $ h = 10 \, \text{m} $ into $ PE=mgh $: $ PE = 2.3\times9.8\times10=225.4 \, \text{J} $ (using $ g = 9.78 \, \text{m/s}^2 $, $ PE=2.3\times9.78\times10\approx224.94 \, \text{J} $).

# Answer: $ \approx 225 \, \text{J} $ (or $ 224.94 \, \text{J} $ to $ 225.4 \, \text{J} $ depending on $ g $ used)

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QUESTION IMAGE

Question

Explanation:

Part a

Step1: Identify the kinematic equation

Step2: Solve for acceleration \( a \)

Step1: Choose the kinematic equation

Step2: Calculate the velocity

Answer:

Part b