Basic Force, Energy, and Work

Definitions

1. Force: In physics, force is defined as a push or pull on an object that causes a change in motion or deformation. It is a vector quantity, which means it has both magnitude (size) and direction. Force is usually measured in units of newtons (N). One newton is equal to one kilogram meter per second squared ($1 N = 1 kg \times m/s^2$).

2. Energy: In physics, energy is a property of objects which can be transferred to other objects or converted into different forms. It is a scalar quantity, which means it has only magnitude and no direction. There are various forms of energy, such as kinetic energy, potential energy, thermal energy, and electromagnetic energy. Energy is usually measured in units of joules (J). One joule is equal to one newton meter ($1 J = 1 N\times m$).

3. Work: In physics, work is defined as the transfer of energy that occurs when a force is applied to an object and causes it to move a distance in the direction of the force. It is a scalar quantity, which means it has only magnitude and no direction. Work is usually calculated by multiplying the force applied to an object by the distance it is moved in the direction of the force, and is measured in units of joules ($J$). One joule is equal to one newton meter ($1 J = 1 N\times m$).

4. Springs: The force exerted by a spring can be described using Hooke’s Law, which states that the force is directly proportional to the displacement of the spring from its equilibrium position. Symbolically, this can be expressed as: $F = -ks$. Where k is the spring constant.

5. Newton’s Three Laws of Motion:

   • Newton’s first law states that if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force.
   • The time rate of change of the momentum of a body is equal in both magnitude and direction to the force imposed on it.
   • When two bodies interact, they apply forces to one another that are equal in magnitude and opposite in direction.

6. The coefficient of kinetic friction is a measure of the frictional force that opposes the motion of an object when it is in motion on a surface. It is a dimensionless constant that describes the relationship between the frictional force and the normal force acting on the object. The coefficient of kinetic friction is denoted by the symbol μk and is determined experimentally by measuring the force required to keep an object moving at a constant speed over a given surface. It varies depending on the nature of the surfaces in contact, the surface roughness, and the conditions under which the motion occurs. The higher the coefficient of kinetic friction, the greater the frictional force and the harder it is to move the object over the surface.

Useful Equations

Kinetic Energy

$E_k = \frac12mv^2$

Spring Potential Energy

$E_p = \frac12ks^2$

Spring Force

$F = -ks$

Frictional force formula

$F_f = \mu \times N$ Where $F_f$ is the frictional force, $μ$ is the coefficient of kinetic friction, and $N$ is the normal force acting on the object.

Work done against friction formula

$W = F_f \times d$ Where $W$ is the work done against friction, $F_f$ is the frictional force, and $d$ is the distance over which the force is applied.

Sliding friction formula

$F_f = \mu mg$

Example Test Questions

A horizontal spring with spring constant y is compressed through a distance y from its equilibrium length. An object of mass m is placed at the end of the spring. The spring is released and returns to its equilibrium length. What is the speed of the object just after it leaves the spring?

Energy in spring $=\frac12ky^2$
Energy to have a speed of $v$ is $\frac12mv^2$
$\therefore \frac12ky^2=\frac12mv^2$
$\implies ky^2=mv^2$
$\implies\frac km y^2=v^2$
$\implies v= \frac{\sqrt k}{\sqrt m} y$

An object of mass $1kg$ is thrown downwards from a height of $20m$. The initial speed of the object is $6ms^{-1}$. The object hits the ground at a speed of $20ms^{-1}$. Assume $g = 10ms^{-2}$. What is the energy transferred from the object to the air as it falls?

Energy lost = Potential energy - Change in kinetic energy
$E_p - \Delta E_k$
$= mgh - (\frac12mv_f^2 - \frac12mv_i^2)$ Where $m$ is the mass, $v_i$ is the initial velocity and $v_f$ is the final velocity.
$= 1\times10\times20 - (\frac12\times1\times20^2 - \frac12\times1\times6^2)$
$= 200 - \frac12(400 - 36)$
$= 200 - 182$
$= 18$

A mass is released from the top of a smooth ramp of height $h$. After leaving the ramp, the mass slides on a rough horizontal surface. The mass comes to rest in a distance $d$. What is the coefficient of dynamic friction between the mass and the horizontal surface?

The potential energy of the mass at the top of the ramp is $mgh$. As the mass slides down the ramp, it gains kinetic energy, which can be calculated using the equation:

$\frac12 mv^2 = mgh$

where v is the velocity of the mass at the bottom of the ramp. Solving for v, we get:

$v = \sqrt{2gh}$

The work done by the frictional force is equal to the change in kinetic energy of the mass:

Work done by friction = kinetic energy at bottom of the ramp - kinetic energy at the point of stop

The kinetic energy at the point of stop is zero, so we can write:

$W=\frac12 mv^2$

Substituting the value of $v$, we get:

$W = mgh$

We know that the work done by the frictional force is equal to the product of the frictional force and the distance d over which it acts:

$W = F_f \times d$
$W = \mu gh \times d$

$\implies μmgd = mgh$
$\implies μ = h/d$

A cart travels from rest along a horizontal surface with a constant acceleration. What is the variation of the kinetic energy $E_k$ of the cart with its distance $s$ travelled? Air resistance is negligible. What is the relationship of $E_k$ to $s$?

$E_k = \frac12mv^2$

$v$ is $t$, time, multiplied by some constant $k$:
$\implies E_k = \frac12m(kt)^2$
$\implies \frac{2E_k}{m} = k^2t^2$
$\implies \frac{2E_k}{mk^2} = t^2$

$s = \frac12kt^2$
$\implies \frac{2s}k =t^2$

$\therefore \frac{2E_k}{mk^2} = \frac{2s}k$
$\implies E_k = mks$

Since $m$ and $k$ are constants $E_k$ and $s$ have a linear relationship

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