You drop a 2.3 kg textbook to a friend who stands on the ground 9 m below the textbook with outstretched hands 1.50 m above the ground.
What is the speed of the textbook when it reaches the hands?
Ignoring air resistance, use conservation of energy.
mg(H-h) = (1/2)mv²
solve for v.
To determine the speed of the textbook when it reaches the hands, we can use the principle of conservation of energy. The initial potential energy of the textbook is converted into kinetic energy as it falls.
Step 1: Find the potential energy of the textbook at the initial height.
The potential energy (PE) of an object is given by the formula PE = mgh, where m is the mass, g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the height.
In this case, the mass (m) of the textbook is 2.3 kg and the initial height (h) is 9 m.
PE = (2.3 kg)(9.8 m/s²)(9 m)
PE = 200.46 J (to two decimal places)
Step 2: Find the potential energy of the textbook at the final height.
The final height is 1.50 m above the ground, so the potential energy at this height is:
PE = (2.3 kg)(9.8 m/s²)(1.50 m)
PE = 33.66 J (to two decimal places)
Step 3: Calculate the kinetic energy of the textbook at the final height.
The difference in potential energy (ΔPE) between the initial and final heights is converted into kinetic energy (KE) as the textbook falls.
ΔPE = PE_final - PE_initial
ΔPE = 33.66 J - 200.46 J
ΔPE ≈ -166.8 J (negative sign indicates a decrease in potential energy)
Step 4: Convert the change in potential energy to kinetic energy.
The equation for kinetic energy is KE = 1/2 mv², where m is the mass of the object and v is the velocity/speed.
So, -166.8 J = 1/2 (2.3 kg) v²
Step 5: Solve for v².
To solve for v², we need to rearrange the equation:
-166.8 J = 1/2 (2.3 kg) v²
v² = -166.8 J / (1/2 (2.3 kg))
v² = -144.69 (to two decimal places)
Step 6: Take the square root to find v.
Taking the square root of v² gives us the speed (v) of the textbook:
v = √(-144.69)
v ≈ 12.03 m/s (to two decimal places)
Therefore, the speed of the textbook when it reaches the hands is approximately 12.03 m/s.
To find the speed of the textbook when it reaches the hands, we can use the principles of conservation of energy. At the instant the textbook reaches the hands, its gravitational potential energy will be converted entirely to kinetic energy.
First, let's calculate the gravitational potential energy of the textbook at the initial height of 1.50 m above the ground. We can use the formula:
Potential Energy = mass * gravitational acceleration * height
Here are the given values:
Mass of the textbook (m) = 2.3 kg
Height above the ground (h) = 1.50 m
Gravitational acceleration (g) = 9.8 m/s²
Potential Energy = 2.3 kg * 9.8 m/s² * 1.50 m = 33.93 J
Next, using the principle of conservation of energy, this potential energy is converted into kinetic energy at the final height of 9 m above the ground. The formula for kinetic energy is:
Kinetic Energy = 0.5 * mass * velocity^2
Since the initial kinetic energy is zero, we can equate the initial potential energy to the final kinetic energy:
Potential Energy = Kinetic Energy
33.93 J = 0.5 * 2.3 kg * velocity^2
Now, we can solve for the velocity:
velocity^2 = (33.93 J) / (0.5 * 2.3 kg)
velocity^2 = 29.43 m²/s²
Taking the square root of both sides to solve for velocity:
velocity = √(29.43 m²/s²)
Calculating this gives us:
velocity ≈ 5.42 m/s
Therefore, the speed of the textbook when it reaches the hands is approximately 5.42 m/s.