A car with a mass of 100 kg, with initial v = 1 m/s, has 60 Joules of energy added for 50 m. What is the final velocty?
I'm using KE = 1/2 mv^2 to find initial KE. Then add 600 J to this for the final KE. Thus, W = KFf - KEi;
KEf = W + KEi
and KEf = 1/2mv^2
v^2 = (2(KEf))/m
Method looks fine to me but is it 60 or 600 Joules that you add?
The "for 50 m" part of your question confuses me. The distance over which the energy is added makes no difference in the final velocity.
You method is correct, but as Damon noted, you have two different numbers for the added energy.
Oop's, my bust. It's 60 Joules. I believe the 50m was thrown is a curve to throw me off, as as the '60J' = W.
Otherwise it would have been say '60N over 50m = 3000J'.
Thanks guys.
To find the final velocity of the car, we can use the law of conservation of energy. The initial kinetic energy (KEi) of the car can be calculated using the formula KE = 1/2 mv^2, where m is the mass of the car and v is the initial velocity. In this case, the mass is given as 100 kg and the initial velocity is 1 m/s.
KEi = 1/2 * 100 kg * (1 m/s)^2
KEi = 1/2 * 100 kg * 1 m^2/s^2
KEi = 50 Joules
The energy added to the car is given as 60 Joules. To find the final kinetic energy (KEf), we can add this energy to the initial kinetic energy.
KEf = KEi + 60 Joules
KEf = 50 Joules + 60 Joules
KEf = 110 Joules
Now, we can use the formula for kinetic energy to find the final velocity (v).
KEf = 1/2 * 100 kg * v^2
110 Joules = 1/2 * 100 kg * v^2
v^2 = (110 Joules * 2) / 100 kg
v^2 = 220 Joules / 100 kg
v^2 = 2.2 m^2/s^2
To find the final velocity (v), we take the square root of both sides of the equation.
v = sqrt(2.2 m^2/s^2)
v ≈ 1.483 m/s
Therefore, the final velocity of the car is approximately 1.483 m/s.