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 **$98\text{ J}$**.

Let's use energy conservation to solve this problem. The initial potential energy of the block at the top of the incline is converted into kinetic energy at the bottom:

$$
\begin{equation*}
    P_i + K_i = P_f + K_f \, .
\end{equation*}
$$

Here, $P$ represents potential energy and $K$ represents kinetic energy. Initial ($i$) and final ($f$) are the two states of a block. The mass is 2 kg throughout this problem so we will leave it out of the equations for simplicity.

At the top: 

$$
\begin{align*}
    P_i &= mgh \\
    K_i &= 0 \, .
\end{align*}
$$
 
Where $m$ represents mass, and $h$ is height ($5 \text{ m}$). At the bottom: 

$$
\begin{align*}
   P_f &= 0 \\
   K_f &= \frac{1}{2}v^2 \, .  
\end{align*}
$$

Where $v$ represents velocity. 

Plugging into our energy conservation equation, we get: 

$$
\begin{equation*}
     mgh + 0 = 0 + \frac{1}{2}v^2 \, .
\end{equation*}
$$
 
Solving for $v$, we find the speed at the bottom of the incline: 

$$
\begin{align*}
    v &= \sqrt{2 mgh} \\
     &= \sqrt{2 (2 \text{ kg})(9.8 \text{ m/s}^2)(5 \text{ m})}\\
     &\approx 14.0 \text{ m/s} \, .
\end{align*}
$$
 
Now that we have the speed at the bottom, finding kinetic energy is easy: 

$$
\begin{align*}
    K_f &= \frac{1}{2}v^2 \\
     & = \frac{1}{2}(14.0 \text{ m/s})^2\\
      &= 98 \text{ J} \, .
\end{align*}
$$
 
So the block's speed at the bottom is $14.0$ meters per second and its kinetic energy at the bottom is $98$ Joules.

Given a 2 kg block sliding down a frictionless incline, dropping a height of 5 m, we need to calculate the block's speed and kinetic energy at the bottom using energy conservation.

First, we use the principle of energy conservation where the potential energy (PE) at the top converts into kinetic energy (KE) at the bottom. The formulas used are:

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

Since all potential energy is converted into kinetic energy:

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

The mass \( m \) cancels out, simplifying to:

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

Plugging in the given values (\( g = 9.8 \, \text{m/s}^2 \) and \( h =

<think>
Okay, so I need to solve this physics problem about a block sliding down a frictionless incline. Let me start by understanding what's given and what's being asked. The problem says a 2 kg block starts from rest at the top of a frictionless incline and slides down, dropping a height of 5 meters. We have to use energy conservation and g = 9.8 m/s² to find two things: (a) the block's speed at the bottom, and (b) its kinetic energy at the bottom. 

Hmm, energy conservation. That usually means that the potential energy at the top will convert into kinetic energy at the bottom, right? Since there's no friction, all the gravitational potential energy should turn into kinetic energy. Let me recall the formulas. 

Gravitational potential energy (PE) is mass times gravity times height, so PE = mgh. Kinetic energy (KE) is (1/2)mv². Since energy is conserved, the potential energy at the top equals the kinetic energy at the bottom. So, I can set mgh equal to (1/2)mv². 

Let me write that equation down:

mgh = (1/2)mv²

Okay, so if I solve for v, I can cancel out the mass since it's on both sides. That gives me:

gh = (1/2)v²  
Then multiply both sides by 2:  
2gh = v²  
So, v = sqrt(2gh)

Plugging in the numbers:  
g is 9.8 m/s², h is 5 m.  
So, v = sqrt(2 * 9.8 * 5). Let me calculate that step by step. 

First, 2 times 9.8 is 19.6. Then, 19.6 multiplied by 5 is 98. So the square root of 98. Hmm, sqrt(98). Let me see, 10 squared is 100, so sqrt(98) is a bit less than 10. Maybe 9.899? Let me check with a calculator. Wait, maybe 9.9? Let me compute 9.9 squared: 9.9^2 is 98.01. Oh, exactly. So sqrt(98) is approximately 9.899 m/s. So the speed should be about 9.9 m/s. But maybe they want an exact value or should I keep it symbolic? Wait, the problem says to give the final numbers with units, so probably needs a numerical value. Let me confirm the calculation again:

2*9.8 = 19.6  
19.6*5 = 98  
sqrt(98) ≈ 9.899 m/s. Since they use g=9.8, maybe they expect an exact form? Wait, 98 is 49*2, so sqrt(49*2) = 7*sqrt(2). Since sqrt(2) is about 1.414, so 7*1.414 is 9.898. That's the same as before. But maybe the question expects to write it as 7√2? But part (a) is asking for the speed, so probably wants the decimal value. Let me see. Since they specify g=9.8, it might be better to compute with that. Let me check:

Alternatively, if they want an exact answer, but 9.8 is a decimal, so maybe we should use the approximate decimal. I think 9.9 m/s is okay, but let's see. Maybe more precise. Let me compute sqrt(98):

sqrt(98) = 9.89949493661... So rounding to two decimal places would be 9.90 m/s. Hmm. Alternatively, maybe the problem expects it as sqrt(98) m/s? But the question says to use energy conservation and give numbers with units, so probably compute the numerical value. Alright, I'll go with approximately 9.9 m/s.

Now part (b) is the kinetic energy at the bottom. Well, since they already used energy conservation, maybe we can just compute the kinetic energy as mgh, because the potential energy is converted to kinetic. Let me see: potential energy at the top is mgh = 2 kg * 9.8 m/s² *5 m. Let me calculate that. 

2 times 9.8 is 19.6, times 5 is 98. So the kinetic energy at the bottom is 98 Joules. Alternatively, using KE = 1/2 mv², with v= sqrt(2gh), so substituting back, 1/2 * m * 2gh = mgh, so that's the same. So either way, it's 98 Joules. That seems straightforward.

Wait, but let me confirm the calculation again. 2 kg * 9.8 m/s² *5 m. 

2*9.8 = 19.6, 19.6*5: 10*5=50, 9.6*5=48, so 50+48=98. Yes, that's correct. So 98 Joules. 

Hmm, seems like part (a) is 9.9 m/s and part (b) is 98 J. But let me check if I made any mistake in the steps. 

Wait, so the key idea here is that in a frictionless incline, the path doesn't matter, only the vertical drop. So regardless of the angle of the incline, as long as the height is 5 m, the speed at the bottom will be the same as if it had fallen straight down vertically. That's why energy conservation simplifies it so much.

Also, for part (a), the mass cancels out, so the speed doesn't depend on the mass of the block. So even if it were a different mass, as long as height and g are the same, the speed is the same. That makes sense because acceleration due to gravity is independent of mass. 

So, in summary, using energy conservation was the right approach here. Since there's no friction, mechanical energy is conserved, so PE converts to KE. 

Therefore, I think my answers are correct. Let me just recast the steps formally:

(a) Start with energy conservation:

Potential Energy (PE) at top = Kinetic Energy (KE) at bottom.

PE = mgh  
KE = (1/2)mv²

Setting them equal: mgh = (1/2)mv²  
Cancel mass: gh = (1/2)v²  
Solve for v: v = sqrt(2gh)  
Plug in numbers: sqrt(2 *9.8*5) = sqrt(98) ≈ 9.899 m/s ≈ 9.9 m/s

(b) Kinetic energy at the bottom is exactly equal to the potential energy lost, so KE = mgh = 2*9.8*5 = 98 J.

Therefore, the answers are:

a) 9.9 m/s

b) 98 J

Wait, but let me check if part (a) requires more precise decimals. The problem didn't specify, but since g is given as 9.8, maybe we should use two significant figures? Wait,