**Problem 1:**  You are holding a 1 kg rock and standing at the top of a cliff. You drop the rock off the cliff and it falls a distance 10 m. In this problem you can ignore air resistance.

What is the change in the kinetic energy of the rock as it falls that distance?
**MultipleChoice**   :

1) About 100 J.
2) About -10 J.
3) About 0.
4) About 10 J.
5) About -100 J.

**Problem 2:**  A center throws a basketball *m* = 1 kg straight up with an initial speed of *v*0 = ***9.5*** m/s. The ball leaves his hand at shoulder height *h*0 = ***1.9*** m. Let gravitational potential energy be zero at ground level.

**Part (a)** Give the total mechanical energy of the ball *E* in terms of maximum height *h*m it reaches, the mass *m*, and the gravitational acceleration *g*.
**Expression**   :
***E*** = __________________________________________

Select from the variables below to write your expression. Note that all variables may not be required.
**α**, **β**, **ρ**, **θ**, **d**, **g**, **hm**, **ho**, **i**, **j**, **k**, **m**, **P**, **t**, **vo**

**Part (b)** What is the height, *h*m in meters?
**Numeric**   : A numeric value is expected and not an expression.

### Problem 1: Change in Kinetic Energy of the Rock

We know from the **Work-Energy Theorem** that the change in the kinetic energy (ΔKE) of the rock is equal to the change in its gravitational potential energy as it falls. This is because all of the potential energy (PE) converts to kinetic energy (KE) if we ignore air resistance.

The change in gravitational potential energy is given by:

\[
\Delta PE = mgh
\]

Where:
- \( m \) = 1 kg (mass of the rock),
- \( g \) = 9.8 m/s² (acceleration due to gravity),
- \( h \) = 10 m (height).

Now, substitute these values into the equation:

\[
\Delta PE = (1\, \text{kg}) (9.8\, \text{m/s}^2) (10\, \text{m}) = 98\, \text{J}
\]

Since all the potential energy converts to kinetic energy:

\[
\Delta KE = \Delta PE = 98\, \text{J} \approx 100\, \text{J}
\]

Thus, the correct answer is **1) About 100 J**.

---

### Problem 2: Energy of a Basketball Thrown Upwards

#### Part (a): Total Mechanical Energy Expression

The **total mechanical energy** of the basketball when it reaches its maximum height \( h_m \) is the sum of its gravitational potential energy (PE) at the maximum height and its kinetic energy (KE). Since at the maximum height, the velocity is zero, the kinetic energy is zero, and the total mechanical energy is just the potential energy at the maximum height:

\[
E = mgh_m
\]

Where:
- \( m \) = mass of the ball,
- \( g \) = acceleration due to gravity,
- \( h_m \) = maximum height the ball reaches.

Thus, the expression for the total mechanical energy is:

\[
E = m g h_m
\]

#### Part (b): Maximum Height Calculation

We can use the principle of conservation of mechanical energy. The initial total mechanical energy when the ball leaves the hand is the sum of its initial kinetic energy and its initial gravitational potential energy at height \( h_0 \):

\[
E = \frac{1}{2} m v_0^2 + m g h_0
\]

Substitute the known values:
- \( m = 1 \, \text{kg} \),
- \( v_0 = 9.5 \, \text{m/s} \),
- \( g = 9.8 \, \text{m/s}^2 \),
- \( h_0 = 1.9 \, \text{m} \).

\[
E = \frac{1}{2} (1)(9.5)^2 + (1)(9.8)(1.9)
\]
\[
E = \frac{1}{2} (90.25) + (18.62)
\]
\[
E = 45.125 + 18.62 = 63.745 \, \text{J}
\]

Now, at the maximum height \( h_m \), all the energy is gravitational potential energy:

\[
E = m g h_m
\]

Solving for \( h_m \):

\[
h_m = \frac{E}{mg} = \frac{63.745}{(1)(9.8)} = 6.5 \, \text{m}
\]

Thus, the maximum height \( h_m \) is **6.5 meters**.

---

### Summary of Answers:
1. **Problem 1:** Change in kinetic energy is approximately **100 J**.
2. **Problem 2 (a):** Total mechanical energy \( E = mgh_m \).
3. **Problem 2 (b):** Maximum height \( h_m = 6.5 \, \text{m} \).

Would you like further clarification on any part of the solution? 

Here are five related questions for deeper understanding:
1. How does the kinetic energy change if air resistance is considered?
2. What would be the total energy if the ball had a mass of 2 kg?
3. How would the maximum height change if the initial velocity was doubled?
4. Can the work-energy theorem apply if the object doesn’t fall vertically?
5. What happens to the potential energy when the rock hits the ground?

**Tip:** The conservation of mechanical energy applies only in the absence of external forces like air resistance.

You are holding a 1 kg rock and standing at the top of a cliff. You drop the rock off the cliff and it falls a distance 10 m. In this problem you can ignore air resistance.
What is the change in the kinetic energy of the rock as it falls that distance?
A center throws a basketball m = 1 kg straight up with an initial speed of v0 = 9.5 m/s. The ball leaves his hand at shoulder height h0 = 1.9 m. Let gravitational potential energy be zero at ground level. Part (a) Give the total mechanical energy of the ball E in terms of maximum height hm it reaches, the mass m, and the gravitational acceleration g. Part (b) What is the height, hm in meters?

Solve physics problems related to kinetic and potential energy. Learn how to calculate the change in kinetic energy for a 1 kg rock falling from a 10 m height and determine the total mechanical energy and maximum height for a basketball thrown straight up. Suitable for students in Grades 9-11.

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