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Chapter 7: Problem 19

A car weighing \(1400 \mathrm{~kg}\) is moving at a speed of \(54 \mathrm{kmh}^{-1}\) up a hill when the motor stops. If it is just able to reach the destination which is at a height of \(10 \mathrm{~m}\) above the point, then the work done against friction (negative of the work done by the friction) is (Take \(g=10 \mathrm{~ms}^{-2}\) ) (a) \(10 \mathrm{~kJ}\) (b) \(15 \mathrm{~kJ}\) (c) \(17.5 \mathrm{~kJ}\) (d) \(25 \mathrm{~kJ}\)

Short Answer

Expert verified
The work done against friction is 17.5 kJ.

Step by step solution

01

Convert the Speed

Convert the speed of the car from km/h to m/s. The car's speed is given as 54 km/h.To convert km/h to m/s, use the formula: \[\text{Speed in } \mathrm{m/s} = \text{Speed in } \mathrm{km/h} \times \frac{5}{18}\]So, the speed is:\[54 \times \frac{5}{18} = 15 \mathrm{~m/s}\]
02

Calculate Initial Kinetic Energy

The kinetic energy (KE) of the car when it starts moving up the hill can be calculated using the formula:\[ KE = \frac{1}{2}mv^2 \]Where:- \(m = 1400 \mathrm{~kg}\) is the mass of the car,- \(v = 15 \mathrm{~m/s}\) is the speed of the car.Substitute the values:\[ KE = \frac{1}{2} \times 1400 \times (15)^2 = 157500 \mathrm{~J} \]
03

Calculate Potential Energy at Height

Calculate the gravitational potential energy (PE) at the height of 10 meters using the formula:\[ PE = mgh \]Where:- \(m = 1400 \mathrm{~kg}\),- \(g = 10 \mathrm{~ms}^{-2}\),- \(h = 10 \mathrm{~m}\).Substitute the values:\[ PE = 1400 \times 10 \times 10 = 140000 \mathrm{~J} \]
04

Determine the Work Done Against Friction

The work done against friction is the difference between the initial kinetic energy and the gravitational potential energy at the height. It can be calculated as:\[ \text{Work Done (against friction)} = KE - PE \]Substitute the values from Step 2 and Step 3:\[ 157500 - 140000 = 17500 \mathrm{~J} \]Convert \(17500 \mathrm{~J}\) to \(17.5 \mathrm{~kJ}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. It depends on two main factors: the mass of the object (\( m \)) and the speed or velocity (\( v \)) at which the object is moving. The formula for calculating kinetic energy is \( KE = \frac{1}{2}mv^2 \).
  • Here, \( m \) represents the mass of the car, measured in kilograms (kg).
  • \( v \) is the velocity of the car in meters per second (m/s).
In the given problem, a car of mass 1400 kg was moving at a speed of 15 m/s. By plugging these values into the kinetic energy formula, we find that the initial kinetic energy was \( 157500 \text{ J} \) (joules). This energy is crucial because it is the source of energy that the car uses to climb the hill.
Understanding kinetic energy helps us see how much energy is available for a moving object to perform work, such as moving against friction or overcoming gravity.
Gravitational Potential Energy
Gravitational potential energy (GPE) is the energy that an object possesses due to its position in a gravitational field. This type of energy is associated with height, so when an object is lifted to a higher position, its potential energy increases. The formula used to calculate gravitational potential energy is \( PE = mgh \).
  • \( m \) stands for the mass of the object, which is the car in this case, expressed in kilograms (kg).
  • \( g \) is the acceleration due to gravity, typically \( 9.8 \text{ m/s}^2 \), but here approximated to \( 10 \text{ m/s}^2 \).
  • \( h \) signifies the height the object has reached, measured in meters (m).
In the scenario, the car is raised to a height of 10 meters. The gravitational potential energy at this height is \( 140000 \text{ J} \). Understanding gravitational potential energy is essential because it helps us calculate how much energy is stored when an object is elevated from one point to another.
Conversion of Units
Conversion of units is an essential skill in physics, allowing us to express measurements in different scales for consistency in calculations. In this exercise, the car's speed needed to be converted from kilometers per hour (km/h) to meters per second (m/s), a common practice to align with standardized units for calculations involving kinetic energy.
The conversion is done using the factor \( \frac{5}{18} \). Simply, you multiply the speed in km/h by this factor to convert it to m/s. For example, 54 km/h \( \times \frac{5}{18} = 15 \text{ m/s} \).
  • Why do we use \( \frac{5}{18} \)? Because the number 5/18 is derived from the conversion factors of 1000 meters in a kilometer, and 3600 seconds in an hour. This makes calculations consistent and accurate within the metric system.
Thus, mastering unit conversion is crucial for ensuring the accuracy of your calculations and understanding the physical quantities involved in a problem.

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Most popular questions from this chapter

The potential energy function for a particle executing linear SHM is given by \(V(x)=\frac{1}{2} k x^{2}\) where \(k\) is the force constant of the oscillator. For \(k=0.5 \mathrm{~N} / \mathrm{m}\), the graph of \(V(x)\) versus \(x\) is shown in the figure. A particle of total energy \(E\) turns back when it reaches \(x=\pm x_{m}\). If \(V\) and \(K\) indicate the \(\mathrm{PE}\) and \(\mathrm{KE}\) respectively of the particle at \(x=\pm x_{m}\) then which of the following is correct? \(\quad\) [NCERT Exemplar] (a) \(V=0, \bar{K}=\mathrm{E}\) (b) \(V=\mathrm{E}, \bar{K}=0\) (c) \(V

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