Work & impulse estimate A car, initially stopped at a stoplight on a level road

Question

Work & impulse estimate

A car, initially stopped at a stoplight on a level road in a residential area, accelerates up to the speed limit.
Ignoring air resistance, identify the three forces acting on the car during the interval described above; use appropriate symbols for each force
Make reasonable assumptions about the car's mass and motion; justify each assumption
Use your assumptions to calculate the work done on the car by each of the three forces
Use your assumptions to calculate the magnitude of the impulse delivered to the car by each of the three forces

For an extra challenge, use the smallest possible number of assumptions. This exercise can be done with just two!
Show and explain your work, and as usual for estimates, use no more than two SF.

Explanation / Answer

The three force, ignoring air resistance, are gravity, normal force, and friction of tires on road. The symbols are usually, Ffs ( this kind of friction is static, not kinetic ), Fn ( normal force ) and W ( weight of vehicle )

An approximation for a cars mass is around 2000 kg, or 2 metric tons.

The motion, for a typical car, can go 0 mph to 60 mph in around 7 seconds. These are the only two assumptions we will use.

Work = Force times distance. Lets find the distance. But, we need to convert 60 mph to m/s, so:

( 60 mile/hour )( 1 hour / 3600 sec )( 1609 m / 1 mile ) = 27 m/s ( 2 significant figures ).

Lets use vf = vi + at to find the typical acceleration.

27 m/s = 0 m/s + a( 7 sec )

a = 3.9 m/s2

Now, lets find the distance

d = viT + ( 1/2 )aT2

d = 0 + ( 1/2 )( 3.9 m/s2)( 7 sec )2

d = 96 m

Now, lets find the force:

F=ma, so F = ( 2000 kg )( 3.9 m/s2 )

F = 7800 N.

Now, finally, the work: W = Fd

W = ( 7800 N )( 96 m ) = 750,000 J

( we could have also used kinetic energy, where Energy = ( 1/2)mv2, which still equals 750,000 J )

The other two forces are vertical, and since there is no vertical displacement, the other forces do zero work in the vertical directions.

The impulse is found using J = Ft OR p = mv, which will both work.

J = ( 7800 N )( 7 sec ) OR p = ( 2000 kg )(27 m/s) = 55000 kg m/s

The The normal force and the wieght cancel out, so while one provides an upwards impulse on the car, the other provides a downwards impulse, and they cancel out.

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