Example 3:
Susan rides her bicycle at a constant speed of
on a horizontal surface. She reaches
point D at the bottom of a slope that has a height
of and a length of 8 m . She freewheels to
the top of the slope, without pedalling, and
experiences a constant frictional force of 20 N
The total mass of Susan and the bicycle is 60 kg .

Identify all of the forces doing work on Susan and calculate the work don
by each of the forces.

Calculate the velocity at which Susan reaches the top of the slope.

Was the mechanical energy conserved while she moved up the slope?

Ask by Chavez Weaver. in South Africa

Mar 05,2025

Question

Example 3:
Susan rides her bicycle at a constant speed of
on a horizontal surface. She reaches
point D at the bottom of a slope that has a height
of and a length of 8 m . She freewheels to
the top of the slope, without pedalling, and
experiences a constant frictional force of 20 N
The total mass of Susan and the bicycle is 60 kg .

Identify all of the forces doing work on Susan and calculate the work don
by each of the forces.

Calculate the velocity at which Susan reaches the top of the slope.

Was the mechanical energy conserved while she moved up the slope?

Ask by Chavez Weaver. in South Africa

Mar 05,2025

UpStudy AI · Accepted Answer

To solve this problem, we will follow the steps outlined in the prompt.

### Step 1: Identify Forces and Calculate Work Done

**Forces acting on Susan:**
1. **Gravitational Force (Weight)**: This force acts downward and can be calculated as:
   $$
   F_g = m \cdot g
   $$
   where $$ m = 60 \, \text{kg} $$ and $$ g = 9.81 \, \text{m/s}^2 $$.

2. **Frictional Force**: This force acts opposite to the direction of motion and is given as $$ F_f = 20 \, \text{N} $$.

**Calculating Work Done by Each Force:**

1. **Work Done by Gravitational Force**:
   The work done by gravity when moving up the slope can be calculated using:
   $$
   W_g = F_g \cdot h
   $$
   where $$ h = 1.5 \, \text{m} $$.

2. **Work Done by Frictional Force**:
   The work done by friction is given by:
   $$
   W_f = -F_f \cdot d
   $$
   where $$ d = 8 \, \text{m} $$ (the length of the slope).

Now, let's calculate these values.

### Step 2: Calculate Velocity at the Top of the Slope

To find the velocity at the top of the slope, we can use the work-energy principle, which states that the work done on an object is equal to the change in its kinetic energy.

The initial kinetic energy $$ KE_i $$ when Susan reaches point D is:
$$
KE_i = \frac{1}{2} m v^2
$$
where $$ v = 10 \, \text{m/s} $$.

The final kinetic energy $$ KE_f $$ at the top of the slope can be expressed as:
$$
KE_f = KE_i + W_g + W_f
$$

The final velocity $$ v_f $$ can be calculated from $$ KE_f $$:
$$
KE_f = \frac{1}{2} m v_f^2
$$

### Step 3: Check for Conservation of Mechanical Energy

Mechanical energy is conserved if the total work done by non-conservative forces (like friction) is zero. We will check if the work done by friction and gravity results in a change in mechanical energy.

Now, let's perform the calculations.

#### Calculating Forces and Work Done

1. **Gravitational Force**:
   $$
   F_g = 60 \, \text{kg} \cdot 9.81 \, \text{m/s}^2
   $$

2. **Work Done by Gravity**:
   $$
   W_g = F_g \cdot h
   $$

3. **Work Done by Friction**:
   $$
   W_f = -20 \, \text{N} \cdot 8 \, \text{m}
   $$

4. **Initial Kinetic Energy**:
   $$
   KE_i = \frac{1}{2} \cdot 60 \, \text{kg} \cdot (10 \, \text{m/s})^2
   $$

5. **Final Kinetic Energy**:
   $$
   KE_f = KE_i + W_g + W_f
   $$

6. **Final Velocity**:
   $$
   KE_f = \frac{1}{2} \cdot 60 \, \text{kg} \cdot v_f^2
   $$

Let's calculate these values.
Calculate the value by following steps:
- step0: Calculate:
  $$60\times 9.81\times 1.5$$
- step1: Multiply the terms:
  $$588.6\times 1.5$$
- step2: Multiply the numbers:
  $$882.9$$
Calculate or simplify the expression $$ 0.5 * 60 * (10^2) $$.
Calculate the value by following steps:
- step0: Calculate:
  $$0.5\times 60\times 10^{2}$$
- step1: Multiply the terms:
  $$30\times 10^{2}$$
- step2: Evaluate the power:
  $$30\times 100$$
- step3: Multiply:
  $$3000$$
Calculate or simplify the expression $$ 60 * 9.81 $$.
Calculate the value by following steps:
- step0: Calculate:
  $$60\times 9.81$$
- step1: Multiply the numbers:
  $$588.6$$
Calculate or simplify the expression $$ -20 * 8 $$.
Calculate the value by following steps:
- step0: Calculate:
  $$-20\times 8$$
- step1: Multiply the numbers:
  $$-160$$
Calculate or simplify the expression $$ 3000 + 882.9 - 160 $$.
Calculate the value by following steps:
- step0: Calculate:
  $$3000+882.9-160$$
- step1: Calculate:
  $$3722.9$$
Calculate or simplify the expression $$ 3722.9 = 0.5 * 60 * v_f^2 $$.
Solve the quadratic equation by following steps:
- step0: Solve using square roots:
  $$3722.9=0.5\times 60v_{f}^{2}$$
- step1: Multiply the terms:
  $$3722.9=30v_{f}^{2}$$
- step2: Convert the expressions:
  $$\frac{37229}{10}=30v_{f}^{2}$$
- step3: Swap the sides:
  $$30v_{f}^{2}=\frac{37229}{10}$$
- step4: Divide both sides:
  $$\frac{30v_{f}^{2}}{30}=\frac{\frac{37229}{10}}{30}$$
- step5: Divide the numbers:
  $$v_{f}^{2}=\frac{37229}{300}$$
- step6: Simplify the expression:
  $$v_{f}=\pm \sqrt{\frac{37229}{300}}$$
- step7: Simplify the expression:
  $$v_{f}=\pm \frac{\sqrt{111687}}{30}$$
- step8: Separate into possible cases:
  $$\begin{align}&v_{f}=\frac{\sqrt{111687}}{30}\&v_{f}=-\frac{\sqrt{111687}}{30}\end{align}$$
- step9: Rewrite:
  $$v_{f1}=-\frac{\sqrt{111687}}{30},v_{f2}=\frac{\sqrt{111687}}{30}$$
### Step 1: Forces and Work Done

1. **Gravitational Force**:
   $$
   F_g = 60 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 588.6 \, \text{N}
   $$

2. **Work Done by Gravity**:
   $$
   W_g = F_g \cdot h = 588.6 \, \text{N} \cdot 1.5 \, \text{m} = 882.9 \, \text{J}
   $$

3. **Work Done by Friction**:
   $$
   W_f = -20 \, \text{N} \cdot 8 \, \text{m} = -160 \, \text{J}
   $$

### Step 2: Initial Kinetic Energy

4. **Initial Kinetic Energy**:
   $$
   KE_i = \frac{1}{2} \cdot 60 \, \text{kg} \cdot (10 \, \text{m/s})^2 = 3000 \, \text{J}
   $$

### Step 3: Final Kinetic Energy

5. **Final Kinetic Energy**:
   $$
   KE_f = KE_i + W_g + W_f = 3000 \, \text{J} + 882.9 \, \text{J} - 160 \, \text{J} = 3722.9 \, \text{J}
   $$

### Step 4: Final Velocity

6. **Final Velocity**:
   We set the final kinetic energy equal to the expression for kinetic energy:
   $$
   3722.9 = \frac{1}{2} \cdot 60 \cdot v_f^2
   $$
   Solving for $$ v_f $$:
   $$
   v_f = \frac{\sqrt{111687}}{30} \approx 6.12 \, \text{m/s}
   $$

### Step 5: Conservation of Mechanical Energy

To check if mechanical energy was conserved, we note that the work done by friction is negative, indicating energy was lost to friction. Therefore, mechanical energy was not conserved as some energy was converted to heat due to friction.

### Summary of Results
1. **Work Done by Gravity**: $$ 882.9 \, \text{J} $$
2. **Work Done by Friction**: $$ -160 \, \text{J} $$
3. **Final Velocity at the Top of the Slope**: $$ \approx 6.12 \, \text{m/s} $$
4. **Mechanical Energy Conservation**: No, mechanical energy was not conserved due to the work done by friction.
**Summary:** - **Work Done by Gravity**: 882.9 J - **Work Done by Friction**: -160 J - **Final Velocity at the Top**: ≈6.12 m/s - **Mechanical Energy Conservation**: No, energy was lost to friction.

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