Friday, February 22, 2019
Unizor - Physics4Teens - Energy - Problems on Kinetic Energy
Notes to a video lecture on http://www.unizor.com
Problems on Kinetic Energy
A car engine accelerates a car of mass m from the state at rest to some maximum speed during the time T with constant acceleration a along a straight line.
Ignore loss of mass due to burning fuel.
What is the kinetic energy Ekinof a car at the end of this period of acceleration and what is the work W performed by the car engine during this time?
Vend = a·T
Ekin = m·V²end /2 = m·a²·T²/2
W = F·S = (m·a) · (a·T²/2) =
W = Ekin = m·a²·T²/2
A driver of a car, going with speed V, sees an obstacle and strongly presses on breaks, so its wheels stop spinning. The car slows down to complete stop by the force of friction only at a distance S from the spot when it started to break.
What is the coefficient of kinetic friction μ of car's wheels against the ground?
Assume, the free fall acceleration is g.
Let m be an unknown mass of the car.
Let W be a work of the force of friction to stop the car.
Let Ekin be a kinetic energy of the car in the beginning of breaking. The ending kinetic energy is zero, since the speed at the end is zero. So, the initial kinetic energy should be equal to the work required to stop the car.
Let F be a constant force of friction equal to the car's weight multiplied by a coefficient of kinetic friction.
Ekin = m·V²/2 = W
W = F·S
F = W/S = m·g·μ
μ = W/(m·g·S) = m·V²/(2·m·g·S)
μ = V²/(2·g·S)
Notice independence of the car's mass.
A point-object A of mass m is freely rotating on a thread of a length L that forms an angle φwith vertical.
What is its kinetic energy?
T is a force of tension of the thread. We can represent it as a sum of two components - Tvthat balances the weight P and Th that keeps an object on a circular trajectory (centripetal force).
Tv = m·g
Th = Tv·tan(φ) = m·g·tan(φ)
This centripetal force causes centripetal acceleration
a = V²/R
(see a chapter Mechanics - Superposition of Forces - Resultant Forces - Example 2 in this course)
where V is a linear speed of an object and R is a radius of a circular trajectory, which is equal to L·sin(φ).
Th = m·a = m·V²/R
m·g·tan(φ) = m·V²/R
Kinetic energy is
E = m·V²/2 =
= m·g·tan(φ)·R/2 =
An ideal pendulum of mass mwith a thread length L performs small harmonic oscillations with its angle of deviation from the vertical φ conforming to the equation
φ(t) = φ0·cos(√g/L ·t)
where φ0 is the initial deviation from the vertical, g - free fall acceleration and t - time.
What is its kinetic energy at the lowest point of its trajectory?
We recommend to refresh the properties of a pendulum in the "Mechanics" part of this course in the chapter "Pendulum, Spring".
Let's find the period of oscillations.
Since function y=cos(x) has a period 2π, function y=cos(k·x)has period 2π/k.
In our case the period of oscillation T is equal to
T = 2π·√L/g
The angular velocity is a derivative of the angle of deviation φ(t):
ω(t) = dφ(t)/dt =
= −φ0·√g/L ·sin(√g/L ·t)
At the lowest point of a trajectory the time equals to 1/4 of a period.
Therefore, we can calculate the angular velocity at this point by substituting
t = T/4 = (1/4)·2π·√L/g =
which gives the angular speed at this point as
= −φ0·√g/L ·sin(√g/L ·T/4) =
= −φ0·√g/L ·
·sin(√g/L ·(π/2)·√L/g ) =
The linear speed of a rotating object equals to its angular speed, multiplied by a radius.
Therefore, linear speed V at the lowest point equals to
V = L·ω(T/4) =
The kinetic energy at this point is
E = m·V²/2 = m·g·L·φ0²/2
The following equation for small harmonic oscillation used in this problem
φ(t) = φ0·cos(√g/L ·t)
was obtained in "Mechanics" part of this course in chapter "Pendulum, Spring" by simplification of the differential equation that described the motion of the pendulum. In particular, it was assumed that for small oscillations the angle of deviation from the vertical φand its sine sin(φ) are approximately equal. The exact value of the kinetic energy of a pendulum at its lowest point can be derived using its potential energy at the top position, which we will address in the next chapter, and which is equal to 2m·g·L·sin²(φ0/2). If in this expression we replace sin(φ0)with φ0 (which is acceptable for small angles), we will get the formula for kinetic energy we obtained above.