How much work must be done to stop a 1150-kg car traveling at 145 km/h?
I will be happy to critique your thinking.
The car does the work equal to the initial kinetic energy. Unless it hits a spring, or it is a hybrid car, the work done is converted to heat.
convert 145 km/h into m/s
145 × 1000 ÷ 3600= 40.3
then use this equation:
W=Fd
or use
KE= 1/2 mv^2
depending on what is given
Vmch
Gjf
To calculate the work done to stop a car, we need to know the initial velocity, final velocity, and mass of the car. In this case, the initial velocity is 145 km/h, final velocity is 0 m/s (since we're stopping the car), and the mass is 1150 kg.
First, we need to convert the initial velocity from km/h to m/s. To do this, you can use the conversion factor: 1 km/h = 0.2778 m/s.
145 km/h = 145 × 0.2778 m/s = 40.278 m/s (rounded to three decimal places).
Now, we can proceed to calculate the work done using the work-energy principle:
Work = (1/2) × mass × (final velocity)^2 - (1/2) × mass × (initial velocity)^2
Work = (1/2) × 1150 kg × (0 m/s)^2 - (1/2) × 1150 kg × (40.278 m/s)^2
Work = 0 J - (1/2) × 1150 kg × (40.278 m/s)^2
Work = -0 J - (1/2) × 1150 kg × (1622.689 m^2/s^2)
Work ≈ -1,177,221 J (rounded to the nearest whole number)
So, approximately 1,177,221 joules of work must be done to stop the 1150-kg car traveling at 145 km/h.