R46 · Energy conservation on a frictionless incline

## Step 1: Calculate the potential energy of the block at the top
The potential energy (PE) of an object is given by PE = m * g * h, where m is the mass of the object, g is the acceleration due to gravity, and h is the height above the ground. In this case, m = 2 kg, g = 9.8 m/s^2, and h = 5 m.

## Step 2: Plug in values for potential energy
PE = (2 kg) * (9.8 m/s^2) * (5 m)

## Step 3: Perform the calculation for potential energy
PE = 98 J

## Step 4: Since the system is frictionless, all potential energy will convert to kinetic energy at the bottom
The total energy of the block remains constant as it slides down the incline. Initially, this energy is in the form of potential energy (PE), and at the bottom, it is converted into kinetic energy (KE).

## Step 5: Set up the equation for energy conservation
Since all the initial PE will convert to KE at the bottom, we have: KE = PE

## Step 6: Since KE = 1/2 * m * v^2 and we found that PE = 98 J, set them equal to each other to find v
1/2 * (2 kg) * v^2 = 98 J

## Step 7: Solve for v
v^2 = 98 J / (1/2 * 2 kg)
v^2 = 49 m^2/s^2
v = sqrt(49 m^2/s^2)

## Step 8: Calculate the speed at the bottom of the incline
v = 7 m/s

## Step 9: Calculate the kinetic energy at the bottom using v and the formula KE = 1/2 * m * v^2
KE = (1/2) * (2 kg) * (7 m/s)^2

## Step 10: Perform the calculation for kinetic energy
KE = 49 J

The final answer is: $\boxed{49}$

Given a 2 kg block starting from rest at the top of a frictionless incline and sliding down to the bottom, dropping a height of 5 meters, we use energy conservation to find the speed and kinetic energy at the bottom.

(a) Using energy conservation, the potential energy at the top is converted into kinetic energy at the bottom. The formula for speed is derived from equating potential energy (PE) to kinetic energy (KE):

\[
mgh = \frac{1}{2}mv^2
\]

Canceling mass \( m \) from both sides and solving for \( v \):

\[
gh = \frac{1}{2}v^2 \implies 2gh = v^2 \implies v = \sqrt{2gh}
\]

Substituting \( g = 9

To solve this problem using energy conservation, we'll apply the principle that the total mechanical energy of the system remains constant if there are no non-conservative forces acting on it.

(a) **Calculating the block's speed at the bottom:**

1. **Initial Potential Energy (PE):**
   The initial potential energy is given by \( PE = mgh \), where \( m \) is the mass of the block, \( g \) is the acceleration due to gravity, and \( h \) is the height.

   \[
   PE = 2 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 5 \, \text{m}
   \]

   \[
   PE = 98 \, \text{J}
   \]

2. **Final Kinetic Energy (KE):**
   At the bottom of the incline, all the potential energy has been converted into kinetic energy.

   \[
   KE = \frac{1}{2} m v^2
   \]

   By conservation of energy:

   \[
   PE = KE
   \]

   \[
   98 \, \text{J} = \frac{1}{2} \times 2 \, \text{kg} \times v^2
   \]

3. **Solving for \( v \):**

   \[
   98 = v^2
   \]

   \[
   v = \sqrt{98}
   \]

   \[
   v \approx 9.9 \, \text{m/s}
   \]

(b) **Calculating the block's kinetic energy at the bottom:**

The kinetic energy at the bottom is equal to the initial potential energy since all of it has been converted into kinetic energy.

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

### Summary:
(a) The block's speed at the bottom is approximately \( 9.9 \, \text{m/s} \).
(b) The block's kinetic energy at the bottom is \( 98 \, \text{J} \).

This problem requires the application of the principle of **Conservation of Energy**, as friction is ignored. The total initial mechanical energy ($E_{initial}$) must equal the total final mechanical energy ($E_{final}$).

**Given values:**
*   Mass ($m$) = $2 \text{ kg}$
*   Initial height ($h_i$) = $5 \text{ m}$
*   Final height ($h_f$) = $0 \text{ m}$ (at the bottom)
*   Initial speed ($v_i$) = $0 \text{ m/s}$
*   Gravity ($g$) = $9.8 \text{ m/s}^2$

### 1. Energy Conservation Setup

$$E_{initial} = E_{final}$$
$$\text{KE}_i + \text{PE}_i = \text{KE}_f + \text{PE}_f$$

Where:
*   $\text{KE} = \frac{1}{2} m v^2$
*   $\text{PE} = m g h$

Substituting the known values (and noting that $v_i=0$, so $\text{KE}_i = 0$; and $h_f=0$, so $\text{PE}_f = 0$):

$$0 + m g h_i = \frac{1}{2} m v^2 + 0$$
$$m g h_i = \frac{1}{2} m v^2$$

---

### (a) Calculate the block's speed at the bottom ($v$)

We solve the energy equation for $v$:

$$\frac{1}{2} m v^2 = m g h_i$$
(The mass $m$ cancels out, meaning the final speed is independent of the object's mass.)

$$v^2 = 2 g h_i$$
$$v = \sqrt{2 g h_i}$$

**Calculation:**
$$v = \sqrt{2 (9.8 \text{ m/s}^2) (5 \text{ m})}$$
$$v = \sqrt{98 \text{ m}^2/\text{s}^2}$$
$$v = 9.90 \text{ m/s}$$

---

### (b) Calculate its kinetic energy at the bottom ($\text{KE}_f$)

According to conservation of energy, all the potential energy lost must be converted into kinetic energy gained:
$$\text{KE}_f = \text{PE}_{lost} = m g h_i$$

**Calculation:**
$$\text{KE}_f = (2 \text{ kg}) (9.8 \text{ m/s}^2) (5 \text{ m})$$
$$\text{KE}_f = 98 \text{ J}$$

***

### Final Answers

**(a) The block's speed at the bottom:**
$$v

This problem utilizes the principle of **Conservation of Mechanical Energy**, which states that in a system without non-conservative forces (like friction), the total mechanical energy remains constant.

$$E_{\text{initial}} = E_{\text{final}}$$
$$\text{Kinetic } (\text{KE}) + \text{Potential } (\text{PE})$$

### Given Variables:
*   Mass ($m$) = $2 \text{ kg}$
*   Initial Height ($h_i$) = $5 \text{ m}$
*   Initial Speed ($v_i$) = $0 \text{ m/s}$ (starts from rest)
*   Gravity ($g$) = $9.8 \text{ m/s}^2$

### Energy Conservation Setup:
The block starts at height $h$ and ends at a final height of 0.

$$\frac{1}{2} m v_i^2 + m g h_i = \frac{1}{2} m v_f^2 + m g h_f$$

Since $v_i=0$ and $h_f=0$:
$$0 + m g h = \frac{1}{2} m v_f^2 + 0$$
$$m g h = \frac{1}{2} m v_f^2$$

***

### (a) Calculation of the block's speed at the bottom ($v_f$)

We use the derived conservation equation:
$$m g h = \frac{1}{2} m v_f^2$$

Notice that the mass ($m$) cancels out from both sides, meaning the final speed is independent of the block's mass.

$$g h = \frac{1}{2} v_f^2$$
$$v_f^2 = 2 g h$$
$$v_f = \sqrt{2 g h}$$

Substitute the values:
$$v_f = \sqrt{2 (9.8 \text{ m/s}^2) (5 \text{ m})}$$
$$v_f = \sqrt

This problem requires using the principle of conservation of mechanical energy, which states that since the incline is frictionless, the total initial energy ($\text{E}_i$) equals the total final energy ($\text{E}_f$).

$$ \text{E}_i = \text{E}_f $$
$$ \text{KE}_i + \text{PE}_i = \text{KE}_f + \text{PE}_f $$

### Given Values:
*   Mass ($m$): $2 \text{ kg}$
*   Initial speed ($v_i$): $0 \text{ m/s}$ (starts from rest)
*   Drop height ($\Delta h = H$): $5 \text{ m}$
*   Gravity ($g$): $9.8 \text{ m/s}^2$

### 1. Initial Energy Calculation:
*   Initial Kinetic Energy ($\text{KE}_i$): $\frac{1}{2} m v_i^2 = \frac{1}{2} (2 \text{ kg}) (0)^2 = 0 \text{ J}$
*   Initial Potential Energy ($\text{PE}_i$): $m g H = (2 \text{ kg})(9.8 \text{ m/s}^2)(5 \text{ m}) = 98 \text{ J}$
*   **Total Initial Energy $\text{E}_i$**: $0 \text{ J} + 98 \text{ J} = 98 \text{ J}$

### 2. Final Energy Calculation:
*   Final Potential Energy ($\text{PE}_f$): Since the block is at the bottom, $h=0$. $\text{PE}_f = m g (0) = 0 \text{ J}$

To solve this problem, we use the **Principle of Conservation of Mechanical Energy**. Since the incline is frictionless, the total mechanical energy at the top of the incline is equal to the total mechanical energy at the bottom.

**Given Data:**
*   Mass ($m$) = $2\text{ kg}$
*   Height ($h$) = $5\text{ m}$
*   Gravity ($g$) = $9.8\text{ m/s}^2$
*   Initial velocity ($v_i$) = $0\text{ m/s}$ (starts from rest)

---

### (a) Calculate the block's speed at the bottom

At the top, the energy is entirely **Gravitational Potential Energy ($PE$)**. At the bottom, the height is zero, so the energy has been converted entirely into **Kinetic Energy ($KE$)**.

**1. Set up the conservation equation:**
$$PE_{top} = KE_{bottom}$$
$$mgh = \frac{1}{2}mv^2$$

**2. Simplify the equation:**
Notice that mass ($m$) appears on both sides, so it cancels out:
$$gh = \frac{1}{2}v^2$$
$$v^2 = 2gh$$
$$v = \sqrt{2gh}$$

**3. Plug in the values:**
$$v = \sqrt{2 \times 9.8\text{ m/s}^2 \times 5\text{ m}}$$
$$v = \sqrt{98\text{ m}^2/\text{s}^2}$$
$$v \approx 9.90\text{ m/s}$$

**Final Answer (a):** The speed at the bottom is **$9.90\text{ m/s}$**.

---

### (b) Calculate the block's kinetic energy at the bottom

Since energy is conserved, the kinetic energy at the bottom is exactly equal to the potential energy at the top.

**1. Use the Potential Energy formula:**
$$KE_{bottom} = PE_{top} = mgh$$

**2. Plug in the values:**
$$KE = 2\text{ kg} \times 9.8\text{ m/s}^2 \times 5\text{ m}$$
$$KE = 98\text{ J}$$

*(Alternatively, using $KE = \frac{1}{2}mv^2$ with our calculated velocity: $KE = \frac{1}{2} \times 2 \times 98 = 98\text{ J}$)*

**Final Answer (b):** The kinetic energy at the bottom is **$98\text{ J}$**.