Question

b4 - rt06: cars changing velocity—work done
the situations below show before and after
\snapshots\ of a cars velocity. all cars have the same
mass.
rank the work done on the cars to create these
changes in velocity for the same distance traveled

Explanation:

Step1: Recall the work - energy theorem

The work - energy theorem states that \(W=\Delta K = K_{f}-K_{i}\), where \(K=\frac{1}{2}mv^{2}\) and \(m\) is the mass of the car (constant for all cars).

Step2: Calculate the work done for each car

• Car A:

\(K_{iA}=\frac{1}{2}m(10)^{2}= 50m\), \(K_{fA}=\frac{1}{2}m(20)^{2}=200m\)
\(W_{A}=K_{fA}-K_{iA}=200m - 50m=150m\)

• Car B:

\(K_{iB}=\frac{1}{2}m(10)^{2}=50m\), \(K_{fB}=\frac{1}{2}m(0)^{2}=0\)
\(W_{B}=K_{fB}-K_{iB}=0 - 50m=- 50m\) (magnitude is \(50m\))

• Car C:

\(K_{iC}=\frac{1}{2}m(10)^{2}=50m\), \(K_{fC}=\frac{1}{2}m(-10)^{2}=50m\)
\(W_{C}=K_{fC}-K_{iC}=50m - 50m = 0\)

• Car D:

\(K_{iD}=\frac{1}{2}m(20)^{2}=200m\), \(K_{fD}=\frac{1}{2}m(20)^{2}=200m\)
\(W_{D}=K_{fD}-K_{iD}=200m - 200m=0\)

Answer:

\(A>B>C = D\)