Question
1.A body of mass 10kg is pulled a distance of 20m Across a horizontal surface against resistance totaling 40N. If the body moves with uniform velocity ,find the work done againstthe resistance
2.A car of mass 80kgworking at a constant rate of 15000W ascends a hill of sin^-1(1/98) against a constant resistance to motion of 420N . find:
The acceleration of the car up the hill when traveling with a velocity of 10ms^-1
The maximum speed of the car up the hill
The acceleration of the car down the same hill at the instant when its speed is 20ms^-1, resistance remains constant
2.A car of mass 80kgworking at a constant rate of 15000W ascends a hill of sin^-1(1/98) against a constant resistance to motion of 420N . find:
The acceleration of the car up the hill when traveling with a velocity of 10ms^-1
The maximum speed of the car up the hill
The acceleration of the car down the same hill at the instant when its speed is 20ms^-1, resistance remains constant
Answers
GPT-4o mini
Let's tackle each problem step by step:
### Problem 1: Work Done Against Resistance
1. **Given:**
- Mass of the body \( m = 10 \, \text{kg} \)
- Distance \( d = 20 \, \text{m} \)
- Resistance \( F_{\text{resistance}} = 40 \, \text{N} \)
2. **Work Done Against Resistance:**
Work done \( W \) against the resistance can be calculated using the formula:
\[
W = F_{\text{resistance}} \times d
\]
Substituting the values:
\[
W = 40 \, \text{N} \times 20 \, \text{m} = 800 \, \text{J}
\]
Therefore, the work done against the resistance is **800 Joules**.
---
### Problem 2: Car Ascending a Hill
Given:
- Mass of the car \( m = 80 \, \text{kg} \)
- Power \( P = 15000 \, \text{W} \)
- Angle of ascent \( \theta = \sin^{-1}(1/98) \) (This means \( \sin(\theta) = 1/98 \))
- Resistance to motion \( F_{\text{resistance}} = 420 \, \text{N} \)
1. **Acceleration of the car up the hill when traveling with a velocity of 10 m/s:**
**Calculate Gravitational Force Component Up the Hill:**
The force due to gravity acting down the hill:
\[
F_{\text{gravity}} = m \cdot g \cdot \sin(\theta)
\]
Where \( g \approx 9.81 \, \text{m/s}^2 \):
\[
F_{\text{gravity}} = 80 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 \cdot \frac{1}{98}
\]
\[
F_{\text{gravity}} \approx \frac{80 \cdot 9.81}{98} \approx 8 \, \text{N} \quad (\text{It's approximately } 8 \, \text{N} \text{ to simplify})
\]
**Total Resistance:**
The total resistance acting on the car:
\[
F_{\text{total}} = F_{\text{gravity}} + F_{\text{resistance}} = 8 \, \text{N} + 420 \, \text{N} = 428 \, \text{N}
\]
**Force Exerted by the Engine:**
Using the power formula:
\[
P = F \cdot v \implies F = \frac{P}{v}
\]
For \( v = 10 \, \text{m/s} \):
\[
F_{\text{engine}} = \frac{15000 \, \text{W}}{10 \, \text{m/s}} = 1500 \, \text{N}
\]
**Net Force and Acceleration:**
\[
F_{\text{net}} = F_{\text{engine}} - F_{\text{total}} = 1500 \, \text{N} - 428 \, \text{N} = 1072 \, \text{N}
\]
Using Newton's second law:
\[
a = \frac{F_{\text{net}}}{m} = \frac{1072 \, \text{N}}{80 \, \text{kg}} \approx 13.4 \, \text{m/s}^2
\]
2. **Maximum Speed of the Car Up the Hill:**
At maximum speed, the net force will be zero (engine force equals total resistance). Therefore,
\[
F_{\text{engine}} = F_{\text{total}} \Rightarrow 1500 \, \text{N} = 428 \, \text{N} + 80 \cdot 9.81 \cdot \frac{1}{98}
\]
The maximum speed occurs when all the power maintains speed against resistance:
From Power formula:
\[
P = F_{\text{total}} \cdot v_{max} \implies v_{max} = \frac{P}{F_{\text{total}}}
\]
Where \( F_{\text{total}} = 428 N \).
\[
v_{max} = \frac{15000\, W}{428 \, N} \approx 35.06 \, \text{m/s}
\]
3. **Acceleration of the Car Down the Hill at 20 m/s:**
In this case:
At \( v = 20 \, \text{m/s} \), the engine force does not work against the movement due to the car's descent; thus, it can be expressed as:
\[
F_{\text{engine}} = 0, \quad F_{\text{gravity} (up)} + F_{\text{resistance}} = m \cdot a
\]
Then,
\[
F_{\text{net}} = F_{\text{gravity}} - F_{\text{resistance}} = 8 \, N - 420 \, N = -412 \, N
\]
This indicates that overall there will be negative acceleration (deceleration), thus
Final acceleration:
\[
a_{down} = \frac{F_{\text{net}}}{80 \, kg} = \frac{-412 \, N}{80 \, kg} = -5.15 \, \text{m/s}^2
\]
### Summary of Answers:
1. Work Done Against Resistance: **800 Joules**
2. Acceleration of the car up the hill at 10 m/s: **13.4 m/s²**
3. Maximum Speed of the car up the hill: **35.06 m/s**
4. Acceleration of the car down the hill at 20 m/s: **-5.15 m/s²** (indicating deceleration).
### Problem 1: Work Done Against Resistance
1. **Given:**
- Mass of the body \( m = 10 \, \text{kg} \)
- Distance \( d = 20 \, \text{m} \)
- Resistance \( F_{\text{resistance}} = 40 \, \text{N} \)
2. **Work Done Against Resistance:**
Work done \( W \) against the resistance can be calculated using the formula:
\[
W = F_{\text{resistance}} \times d
\]
Substituting the values:
\[
W = 40 \, \text{N} \times 20 \, \text{m} = 800 \, \text{J}
\]
Therefore, the work done against the resistance is **800 Joules**.
---
### Problem 2: Car Ascending a Hill
Given:
- Mass of the car \( m = 80 \, \text{kg} \)
- Power \( P = 15000 \, \text{W} \)
- Angle of ascent \( \theta = \sin^{-1}(1/98) \) (This means \( \sin(\theta) = 1/98 \))
- Resistance to motion \( F_{\text{resistance}} = 420 \, \text{N} \)
1. **Acceleration of the car up the hill when traveling with a velocity of 10 m/s:**
**Calculate Gravitational Force Component Up the Hill:**
The force due to gravity acting down the hill:
\[
F_{\text{gravity}} = m \cdot g \cdot \sin(\theta)
\]
Where \( g \approx 9.81 \, \text{m/s}^2 \):
\[
F_{\text{gravity}} = 80 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 \cdot \frac{1}{98}
\]
\[
F_{\text{gravity}} \approx \frac{80 \cdot 9.81}{98} \approx 8 \, \text{N} \quad (\text{It's approximately } 8 \, \text{N} \text{ to simplify})
\]
**Total Resistance:**
The total resistance acting on the car:
\[
F_{\text{total}} = F_{\text{gravity}} + F_{\text{resistance}} = 8 \, \text{N} + 420 \, \text{N} = 428 \, \text{N}
\]
**Force Exerted by the Engine:**
Using the power formula:
\[
P = F \cdot v \implies F = \frac{P}{v}
\]
For \( v = 10 \, \text{m/s} \):
\[
F_{\text{engine}} = \frac{15000 \, \text{W}}{10 \, \text{m/s}} = 1500 \, \text{N}
\]
**Net Force and Acceleration:**
\[
F_{\text{net}} = F_{\text{engine}} - F_{\text{total}} = 1500 \, \text{N} - 428 \, \text{N} = 1072 \, \text{N}
\]
Using Newton's second law:
\[
a = \frac{F_{\text{net}}}{m} = \frac{1072 \, \text{N}}{80 \, \text{kg}} \approx 13.4 \, \text{m/s}^2
\]
2. **Maximum Speed of the Car Up the Hill:**
At maximum speed, the net force will be zero (engine force equals total resistance). Therefore,
\[
F_{\text{engine}} = F_{\text{total}} \Rightarrow 1500 \, \text{N} = 428 \, \text{N} + 80 \cdot 9.81 \cdot \frac{1}{98}
\]
The maximum speed occurs when all the power maintains speed against resistance:
From Power formula:
\[
P = F_{\text{total}} \cdot v_{max} \implies v_{max} = \frac{P}{F_{\text{total}}}
\]
Where \( F_{\text{total}} = 428 N \).
\[
v_{max} = \frac{15000\, W}{428 \, N} \approx 35.06 \, \text{m/s}
\]
3. **Acceleration of the Car Down the Hill at 20 m/s:**
In this case:
At \( v = 20 \, \text{m/s} \), the engine force does not work against the movement due to the car's descent; thus, it can be expressed as:
\[
F_{\text{engine}} = 0, \quad F_{\text{gravity} (up)} + F_{\text{resistance}} = m \cdot a
\]
Then,
\[
F_{\text{net}} = F_{\text{gravity}} - F_{\text{resistance}} = 8 \, N - 420 \, N = -412 \, N
\]
This indicates that overall there will be negative acceleration (deceleration), thus
Final acceleration:
\[
a_{down} = \frac{F_{\text{net}}}{80 \, kg} = \frac{-412 \, N}{80 \, kg} = -5.15 \, \text{m/s}^2
\]
### Summary of Answers:
1. Work Done Against Resistance: **800 Joules**
2. Acceleration of the car up the hill at 10 m/s: **13.4 m/s²**
3. Maximum Speed of the car up the hill: **35.06 m/s**
4. Acceleration of the car down the hill at 20 m/s: **-5.15 m/s²** (indicating deceleration).