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Chapter 6: Problem 7

The momentum of a body of mass \(5 \mathrm{~kg}\) is \(10 \mathrm{~kg} \mathrm{~m} / \mathrm{s}\). A force of \(2 \mathrm{~N}\) acts on the body in the direction of motion for \(5 \mathrm{sec}\), the increase in the kinetic energy is: (a) \(15 \mathrm{~J}\) (b) \(50 \mathrm{~J}\) (c) \(30 \mathrm{~J}\) (d) none of these

Short Answer

Expert verified
The increase in kinetic energy is (c) 30 J.

Step by step solution

01

Determine Initial Velocity

The initial momentum of the body is given by \( p = mv = 10 \text{ kg m/s} \). We have the mass \( m = 5 \text{ kg} \). Using \( p = mv \), we find the initial velocity \( v_i = \frac{p}{m} = \frac{10}{5} = 2 \text{ m/s} \).
02

Calculate Initial Kinetic Energy

The initial kinetic energy \( KE_i \) can be calculated using the formula \( KE = \frac{1}{2} mv^2 \). Substituting the known values, \( KE_i = \frac{1}{2} \times 5 \times (2)^2 = 10 \text{ J} \).
03

Determine Final Velocity with Impulse-Momentum Theorem

The impulse from the force is given by \( F \times t = 2 \times 5 = 10 \text{ Ns} \). By the impulse-momentum theorem, this change in momentum \( \Delta p = F \times t = 10 \text{ Ns} \). The final momentum \( p_f = 10 \text{ kg m/s} (\text{initial momentum}) + 10 \text{ Ns} = 20 \text{ kg m/s} \). Thus, the final velocity \( v_f = \frac{p_f}{m} = \frac{20}{5} = 4 \text{ m/s} \).
04

Calculate Final Kinetic Energy

The final kinetic energy \( KE_f \) is given by \( KE_f = \frac{1}{2} mv_f^2 = \frac{1}{2} \times 5 \times (4)^2 = 40 \text{ J} \).
05

Calculate Increase in Kinetic Energy

The increase in kinetic energy is \( \Delta KE = KE_f - KE_i = 40 - 10 = 30 \text{ J} \).

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Key Concepts

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

Momentum
Momentum is a fundamental concept in physics that refers to the quantity of motion an object has. It is determined by the product of an object's mass and its velocity. Mathematically, momentum is represented as:\[ p = mv \]where \( p \) is momentum, \( m \) is mass, and \( v \) is velocity. In the given exercise, a body with a mass of 5 kg has a momentum of 10 kg m/s, which means it is initially moving at a speed of 2 m/s.
  • Momentum is a vector quantity, meaning it has both magnitude and direction.
  • It plays a crucial role in analyzing motion and predicting the outcome of forces acting on objects.
Understanding momentum helps us see how objects will move and interact when subjected to forces or collisions.
Impulse-Momentum Theorem
The impulse-momentum theorem is a concept that links force to the change in an object's momentum. It states that the change in an object's momentum is equal to the impulse applied to it. Impulse is defined as the product of force and the time duration over which the force acts:\[ Impulse = F imes t \]According to the theorem:\[ \Delta p = F \times t \]In the exercise, a force of 2 N acts for 5 seconds, producing an impulse of 10 Ns. This impulse causes the initial momentum of 10 kg m/s to increase, resulting in a final momentum of 20 kg m/s.
  • This theorem highlights how forces acting over time alter an object's momentum.
  • It provides a method to determine the effect of force on motion over a given time.
Using the theorem makes it easier to understand force impacts in various scenarios, including sports, driving, and aerospace applications.
Force and Motion
Force is a push or pull that can cause an object with mass to change its velocity, i.e., to accelerate. It is directly linked to motion through Newton's second law of motion, which states:\[ F = ma \]where \( F \) is the force applied, \( m \) is the mass of the object, and \( a \) is the acceleration produced. In the context of the exercise, a force of 2 N changes the motion of the object over time, increasing its velocity from 2 m/s to 4 m/s.
  • The direction of the applied force determines the direction of the motion change.
  • Force acting in the direction of motion results in acceleration, while the opposing force causes deceleration.
Understanding force and motion is crucial for predicting how objects will behave when forces are applied, forming the foundation of dynamics.
Kinetic Energy Calculation
Kinetic energy is the energy that an object possesses due to its motion. It can be calculated using the formula:\[ KE = \frac{1}{2} mv^2 \]where \( KE \) is kinetic energy, \( m \) is mass, and \( v \) is velocity. In the exercise, the mass of 5 kg moving at an initial speed of 2 m/s has an initial kinetic energy of 10 J. After a force has acted to increase the speed to 4 m/s, the final kinetic energy becomes 40 J.
  • The increase in kinetic energy illustrates how the speed of an object directly influences its kinetic energy level.
  • Kinetic energy, being a scalar quantity, only has magnitude and is positive.
The calculation of kinetic energy is essential in fields such as engineering, automotive industries, and any scenario involving moving bodies.

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Most popular questions from this chapter

The earth's radius is \(R\) and acceleration due to gravity at its surface is \(g\). If a body of mass \(m\) is sent to a height \(h=\frac{R}{5}\) from the earth's surface, the potential energy increases by : (a) \(m g h\) (b) \(\frac{4}{5} m g h\) (c) \(\frac{5}{6} m g h\) (d) \(\frac{6}{7} m g h\)

A coolie \(1.5 \mathrm{~m}\) tall raises a load of \(80 \mathrm{~kg}\) in \(2 \mathrm{sec}\) from the ground to his head and then walks a distance of \(40 \mathrm{~m}\) in another 2 second. The power developed by the coolie is: \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\) (a) \(0.2 \mathrm{~kW}\) (b) \(0.4 \mathrm{~kW}\) (c) \(0.6 \mathrm{~kW}\) (d) \(0.8 \mathrm{~kW}\)

Force \(F\) acts on a body of mass \(1 \mathrm{~kg}\) moving with an initial velocity \(v_{0}\) for \(1 \mathrm{sec}\). Then : (a) distance covered by the body is \(v_{0}+\frac{F}{2}\) (b) final velocity of body is \(\left(v_{0}+F\right)\) (c) momentum of body is increased by \(F\) (d) all of the above

A force \(F=A y^{2}+B y+C\) acts on a body in the \(y\) -direction. The work done by this force during a displacement from \(y=-a\) to \(y=a\) is : (a) \(\frac{2 A a^{3}}{3}\) (b) \(\frac{2 A a^{3}}{3}+2 C a\) (c) \(\frac{2 A a^{3}}{3}+\frac{B a^{2}}{2}+C a\) (d) none of these

Which one of the following units measures energy? (a) kilo-watt-hour (b) \((\mathrm{vol} \mathrm{t})^{2}(\mathrm{sec})^{-1}(\mathrm{ohm})^{-1}\) (c) (pascal) (foot) \(^{-}\) id) none of the above
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