a) This question involves the work-energy theorem, which states that the work of the external forces acting on a system equals the change of the kinetic energy of the system.
In this case, the external force is the force produced by the engine (even though the engine is of course inside the car, there are outside sources, such as an electric battery, that produced the force). The work of this force over the distance of...
a) This question involves the work-energy theorem, which states that the work of the external forces acting on a system equals the change of the kinetic energy of the system.
In this case, the external force is the force produced by the engine (even though the engine is of course inside the car, there are outside sources, such as an electric battery, that produced the force). The work of this force over the distance of five meters, assuming that the car moves in a straight line, will be
`W=F*d = 2000 N * 5 m = 10,000 J` (Joules.)
Then, the change in the car's kinetic energy is
`Delta K = K_f - K_i = 10,000` J.
Since initially the car is at rest, `K_i = 0` and then `K_f = W = 10,000 J` .
The car's kinetic energy at the end of 5 meters is 10,000 Joules.
b) The formula for kinetic energy is
`K = mv^2/2` , where m is the mass of the car and v is the speed of the car. Since we have found the kinetic energy, we can find the speed:
`v = sqrt((2K)/m) = sqrt((2*10,000)/800) = 5 m/s`
The speed of the car at the end of 5 meters is 5 m/s.
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