Question

question 1 (1 point)
which requires less work: increasing a cars speed from 0 mph to 30 mph or from 50
mph to 60 mph?

a) 50 mph to 60 mph
b) 0 mph to 30 mph
c) it is the same in both cases.
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Explanation:

Step1: Recall the work - energy theorem

The work - energy theorem states that \(W=\Delta K = K_{f}-K_{i}\), where \(K=\frac{1}{2}mv^{2}\) (kinetic energy formula).

Step2: Calculate the work for increasing speed from \(v_{i1} = 0\) mph to \(v_{f1}=30\) mph

Let the mass of the car be \(m\). Then \(K_{i1}=\frac{1}{2}m(0)^{2}=0\) and \(K_{f1}=\frac{1}{2}m(30)^{2}=450m\). So \(W_1=K_{f1}-K_{i1}=450m\).

Step3: Calculate the work for increasing speed from \(v_{i2}=50\) mph to \(v_{f2} = 60\) mph

\(K_{i2}=\frac{1}{2}m(50)^{2}=1250m\) and \(K_{f2}=\frac{1}{2}m(60)^{2}=1800m\). Then \(W_2=K_{f2}-K_{i2}=1800m - 1250m=550m\).

Answer:

B. \(0\) mph to \(30\) mph