By what factor is the kinetic energy increased when the speed of a car is increased from 21 km/h to 98 km/h?
The kinetic energy of an object is given by the equation:
KE = (1/2)mv^2
where KE is the kinetic energy, m is the mass of the object, and v is the velocity of the object.
To calculate the factor by which the kinetic energy is increased, we need to compare the kinetic energy at two different velocities.
Let's assume that the mass of the car remains constant. So, the factor by which the kinetic energy is increased can be found by taking the ratio of the kinetic energies at the two velocities:
Factor = KE2 / KE1
where KE1 is the kinetic energy at the initial velocity and KE2 is the kinetic energy at the final velocity.
First, let's convert the velocities from km/h to m/s, as the unit for velocity in the formula is meters per second (m/s).
Initial velocity (v1) = 21 km/h = (21 * 1000) m/ (60 * 60) s ~= 5.83 m/s
Final velocity (v2) = 98 km/h = (98 * 1000) m/ (60 * 60) s ~= 27.22 m/s
Now, plug the values into the formula to find the kinetic energies:
KE1 = (1/2) * m * v1^2
KE2 = (1/2) * m * v2^2
Factor = KE2 / KE1 = [(1/2) * m * v2^2] / [(1/2) * m * v1^2]
Simplifying the equation:
Factor = v2^2 / v1^2
Now, substitute the values of v1 and v2:
Factor = (27.22 m/s)^2 / (5.83 m/s)^2
Calculating:
Factor = 27.22^2 / 5.83^2
Factor = 746.13 / 33.99
Factor ~= 21.95
Therefore, the kinetic energy is increased by a factor of approximately 21.95 when the speed of the car is increased from 21 km/h to 98 km/h.