3) a mountain climber is on the edge of a cliff 200 meters above a lake.
a. how much potential energy does his 25 kg backpack have?
b. if the backpack is accidentally pushed off, how much kinetic energy will the backpack have halfway down?
c. how fast will the backpack be going when it is halfway down?
d. how much kinetic energy will the backpack have right before it hits the water?
e. how fast will the backpack be going right before it hits the water?

Question

Accepted Answer

### Part a
# Explanation:
## Step1: Recall the formula for gravitational potential energy
The formula for gravitational potential energy is $ PE = mgh $, where $ m $ is the mass, $ g $ is the acceleration due to gravity (we'll use $ g = 9.8 \, \text{m/s}^2 $), and $ h $ is the height.
## Step2: Substitute the given values
We have $ m = 25 \, \text{kg} $, $ g = 9.8 \, \text{m/s}^2 $, and $ h = 200 \, \text{m} $. Plugging these into the formula:
$ PE = 25 \times 9.8 \times 200 $
## Step3: Calculate the result
First, calculate $ 25 \times 9.8 = 245 $. Then, $ 245 \times 200 = 49000 \, \text{J} $.
# Answer:
The potential energy of the backpack is $ 49000 \, \text{joules} $ (or $ 4.9 \times 10^4 \, \text{J} $).

### Part b
# Explanation:
## Step1: Understand the conservation of mechanical energy
Assuming no air resistance, the total mechanical energy (potential + kinetic) is conserved. At the top, the kinetic energy $ KE_{top} = 0 $, so the total mechanical energy $ E_{total} = PE_{top} $. Halfway down, the height is $ h_{half} = \frac{200}{2} = 100 \, \text{m} $. The potential energy at halfway is $ PE_{half} = mgh_{half} $, and the kinetic energy $ KE_{half} = E_{total} - PE_{half} $.
## Step2: Calculate $ PE_{top} $ (which is $ E_{total} $)
From part a, $ PE_{top} = 49000 \, \text{J} $, so $ E_{total} = 49000 \, \text{J} $.
## Step3: Calculate $ PE_{half} $
Using $ PE = mgh $ with $ m = 25 \, \text{kg} $, $ g = 9.8 \, \text{m/s}^2 $, and $ h = 100 \, \text{m} $:
$ PE_{half} = 25 \times 9.8 \times 100 = 24500 \, \text{J} $
## Step4: Calculate $ KE_{half} $
$ KE_{half} = E_{total} - PE_{half} = 49000 - 24500 = 24500 \, \text{J} $
# Answer:
The kinetic energy halfway down is $ 24500 \, \text{joules} $ (or $ 2.45 \times 10^4 \, \text{J} $).

### Part c
# Explanation:
## Step1: Recall the formula for kinetic energy
The formula for kinetic energy is $ KE = \frac{1}{2}mv^2 $, where $ m $ is the mass and $ v $ is the velocity. We know $ KE_{half} = 24500 \, \text{J} $ from part b and $ m = 25 \, \text{kg} $. We need to solve for $ v $.
## Step2: Rearrange the kinetic energy formula to solve for $ v $
Starting with $ KE = \frac{1}{2}mv^2 $, multiply both sides by 2: $ 2KE = mv^2 $. Then divide both sides by $ m $: $ v^2 = \frac{2KE}{m} $. Take the square root of both sides: $ v = \sqrt{\frac{2KE}{m}} $.
## Step3: Substitute the values
Substitute $ KE = 24500 \, \text{J} $ and $ m = 25 \, \text{kg} $:
$ v = \sqrt{\frac{2 \times 24500}{25}} $
## Step4: Calculate the result
First, calculate the numerator: $ 2 \times 24500 = 49000 $. Then, $ \frac{49000}{25} = 1960 $. Take the square root of 1960: $ v = \sqrt{1960} \approx 44.27 \, \text{m/s} $ (or we can simplify $ \sqrt{1960} = \sqrt{196 \times 10} = 14\sqrt{10} \approx 44.27 \, \text{m/s} $).
# Answer:
The speed of the backpack halfway down is approximately $ 44.3 \, \text{m/s} $ (or $ 14\sqrt{10} \, \text{m/s} \approx 44.3 \, \text{m/s} $).

### Part d
# Explanation:
## Step1: Use conservation of mechanical energy
Right before hitting the water, the height $ h = 0 $, so the potential energy $ PE = 0 $. Therefore, all the initial potential energy is converted into kinetic energy. So $ KE_{final} = PE_{initial} $.
## Step2: Recall the initial potential energy
From part a, the initial potential energy $ PE_{initial} = 49000 \, \text{J} $. So $ KE_{final} = 49000 \, \text{J} $.
# Answer:
The kinetic energy right before hitting the water is $ 49000 \, \text{joules} $ (or $ 4.9 \times 10^4 \, \text{J} $).

### Part e
# Explanation:
## Step1: Recall the kinetic energy formula
We know $ KE = \frac{1}{2}mv^2 $, and we know $ KE = 49000 \, \text{J} $ (from part d) and $ m = 25 \, \text{kg} $. We need to solve for $ v $.
## Step2: Rearrange the formula to solve for $ v $
Using $ KE = \frac{1}{2}mv^2 $, we can rearrange it as $ v = \sqrt{\frac{2KE}{m}} $ (same as in part c).
## Step3: Substitute the values
Substitute $ KE = 49000 \, \text{J} $ and $ m = 25 \, \text{kg} $:
$ v = \sqrt{\frac{2 \times 49000}{25}} $
## Step4: Calculate the result
First, calculate the numerator: $ 2 \times 49000 = 98000 $. Then, $ \frac{98000}{25} = 3920 $. Take the square root of 3920: $ v = \sqrt{3920} \approx 62.6 \, \text{m/s} $ (or we can simplify $ \sqrt{3920} = \sqrt{49 \times 80} = 7\sqrt{80} = 7 \times 4\sqrt{5} = 28\sqrt{5} \approx 62.6 \, \text{m/s} $).
# Answer:
The speed of the backpack right before hitting the water is approximately $ 62.6 \, \text{m/s} $ (or $ 28\sqrt{5} \, \text{m/s} \approx 62.6 \, \text{m/s} $).

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QUESTION IMAGE

Question

Explanation:

Part a

Step1: Recall the formula for gravitational potential energy

Step2: Substitute the given values

Step3: Calculate the result

Step1: Understand the conservation of mechanical energy

Step2: Calculate \( PE_{top} \) (which is \( E_{total} \))

Step3: Calculate \( PE_{half} \)

Step4: Calculate \( KE_{half} \)

Step1: Recall the formula for kinetic energy

Step2: Rearrange the kinetic energy formula to solve for \( v \)

Step3: Substitute the values

Step4: Calculate the result

Answer:

Part b