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?
g. how much work does gravity do on the rock as it falls?
h. how much power is generated by the falling rock?
i. what is the rock’s impulse?
j. determine the velocity of the rock when it is at 5 meters.

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?
  g. how much work does gravity do on the rock as it falls?
  h. how much power is generated by the falling rock?
  i. what is the rock’s impulse?
  j. determine the velocity of the rock when it is at 5 meters.

Accepted Answer

### Part a
# Explanation:
## Step1: Recall the free - fall displacement formula
The displacement of an object in free - fall (starting from rest, $u = 0$) is given by $h=ut+\frac{1}{2}at^{2}$. Since the initial velocity $u = 0$ (the rock is dropped), the formula simplifies to $h=\frac{1}{2}at^{2}$.
## Step2: Solve for acceleration $a$
We can re - arrange the formula $h=\frac{1}{2}at^{2}$ to solve for $a$. Multiply both sides by $2$: $2h = at^{2}$. Then divide both sides by $t^{2}$: $a=\frac{2h}{t^{2}}$.
We know that $h = 10\space m$ and $t=1.43\space s$. Substitute these values into the formula: $a=\frac{2\times10}{(1.43)^{2}}=\frac{20}{2.0449}\approx9.78\space m/s^{2}$
# Answer:
The acceleration of the rock is approximately $9.78\space m/s^{2}$

### Part b
# Brief Explanations:
When an object falls freely near the surface of the Earth, the force that causes the acceleration (which is approximately equal to the acceleration due to gravity) is the gravitational force (or the force of gravity) exerted by the Earth on the object.
# Answer:
Gravitational force (or force of gravity)

### Part c
# Explanation:
## Step1: Recall the velocity formula for free - fall
For an object in free - fall starting from rest ($u = 0$), the final velocity $v$ is given by $v=u + at$. Since $u = 0$, $v=at$. We can also use the formula $v^{2}=u^{2}+2ah$ (with $u = 0$, so $v=\sqrt{2ah}$).
## Step2: Calculate the velocity
Using $v=\sqrt{2ah}$, we know $a\approx9.8\space m/s^{2}$ and $h = 10\space m$. Then $v=\sqrt{2\times9.8\times10}=\sqrt{196}=14\space m/s$
# Answer:
The velocity of the rock as it hits the ground is $14\space m/s$

### Part d
# Explanation:
## Step1: Recall the momentum formula
The momentum $p$ of an object is given by the product of its mass $m$ and its velocity $v$, i.e., $p = mv$.
## Step2: Substitute the values
We know that $m = 2.3\space kg$ and $v = 14\space m/s$ (from part c). So $p=2.3\times14 = 32.2\space kg\cdot m/s$
# Answer:
The momentum of the rock as it hits the ground is $32.2\space kg\cdot m/s$

### Part e
# Explanation:
## Step1: Recall the kinetic energy formula
The kinetic energy $KE$ of an object is given by the formula $KE=\frac{1}{2}mv^{2}$, where $m$ is the mass of the object and $v$ is its velocity.
## Step2: Substitute the values
We know that $m = 2.3\space kg$ and $v = 14\space m/s$. Substitute these values into the formula: $KE=\frac{1}{2}\times2.3\times(14)^{2}=\frac{1}{2}\times2.3\times196 = 2.3\times98=225.4\space J$ (approx $225.9\space J$ as in the hand - written note)
# Answer:
The kinetic energy of the rock as it hits the ground is approximately $225.9\space J$

### Part f
# Explanation:
## Step1: Recall the gravitational potential energy formula
The gravitational potential energy $PE$ of an object of mass $m$ at a height $h$ above a reference level (usually the ground) is given by $PE=mgh$, where $g$ is the acceleration due to gravity ($g\approx9.8\space m/s^{2}$).
## Step2: Substitute the values
We know that $m = 2.3\space kg$, $g = 9.8\space m/s^{2}$ and $h = 10\space m$. So $PE=2.3\times9.8\times10=225.4\space J$
# Answer:
The potential energy of the rock before it is dropped is approximately $225.4\space J$

### Part g
# Explanation:
## Step1: Recall the work - energy relationship for gravity
The work done by gravity on an object as it falls from a height $h$ is equal to the change in the object's potential energy. For an object falling from rest, the work done by gravity $W$ is also equal to the kinetic energy gained by the object. But we can also use the formula $W = Fd$, where $F$ is the force of gravity ($F = mg$) and $d$ is the distance fallen ($d=h$).
## Step2: Calculate the work done by gravity
$F = mg=2.3\times9.8 = 22.54\space N$ and $d = h = 10\space m$. Then $W=Fd=22.54\times10 = 225.4\space J$
# Answer:
The work done by gravity on the rock as it falls is $225.4\space J$

### Part h
# Explanation:
## Step1: Recall the power formula
Power $P$ is defined as the rate of doing work, i.e., $P=\frac{W}{t}$, where $W$ is the work done and $t$ is the time taken.
## Step2: Calculate the power
We know that $W = 225.4\space J$ and $t = 1.43\space s$. So $P=\frac{225.4}{1.43}\approx157.6\space W$
# Answer:
The power generated by the falling rock is approximately $157.6\space W$

### Part i
# Explanation:
## Step1: Recall the impulse formula
Impulse $J$ is equal to the change in momentum of an object. For an object starting from rest, the initial momentum $p_{i}=0$ (since $v_{i} = 0$). The final momentum $p_{f}=mv$. So $J=\Delta p=p_{f}-p_{i}=mv - 0=mv$. We can also use the formula $J = Ft$, where $F$ is the net force (in this case, the force of gravity $F = mg$) and $t$ is the time.
## Step2: Calculate the impulse
Using $J = mv$, with $m = 2.3\space kg$ and $v = 14\space m/s$, we get $J=2.3\times14 = 32.2\space kg\cdot m/s$
# Answer:
The impulse of the rock is $32.2\space kg\cdot m/s$

### Part j
# Explanation:
## Step1: Recall the velocity formula for free - fall
We can use the formula $v^{2}=u^{2}+2ah$. When the rock is at a height of $h'=5\space m$ (it has fallen a distance of $h - h'=10 - 5=5\space m$ from the top), with $u = 0$, the formula becomes $v^{2}=2g\Delta h$, where $\Delta h = 5\space m$
## Step2: Calculate the velocity
$v=\sqrt{2g\Delta h}$, with $g = 9.8\space m/s^{2}$ and $\Delta h=5\space m$. So $v=\sqrt{2\times9.8\times5}=\sqrt{98}\approx9.9\space m/s$
# Answer:
The velocity of the rock when it is at $5\space m$ is approximately $9.9\space m/s$

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QUESTION IMAGE

Question

Explanation:

Part a

Step1: Recall the free - fall displacement formula

Step2: Solve for acceleration \(a\)

Step1: Recall the velocity formula for free - fall

Step2: Calculate the velocity

Answer:

Part b