Question

question 1 (1 point)
in which of the following scenarios was the most net work done?
o a a net force of 10 n is applied to an object as it is displaced 1 m
o b an objects kinetic energy is increased by 20 j.
o c a 5 kg objects velocity is increased from 1 m/s to 3 m/s.
o d a net force of 5 n is applied to an object as it is displaced 5 m.
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Explanation:

Step1: Calculate work for option a

Use the formula \(W = Fd\). Here \(F = 10\space N\) and \(d=1\space m\).
\(W_a=10\times1 = 10\space J\)

Step2: Use work - energy theorem for option b

By the work - energy theorem \(W=\Delta K\). Given \(\Delta K = 20\space J\), so \(W_b = 20\space J\)

Step3: Use work - energy theorem for option c

The formula for kinetic energy is \(K=\frac{1}{2}mv^{2}\). Initial velocity \(v_1 = 1\space m/s\) and final velocity \(v_2=3\space m/s\), \(m = 5\space kg\)
\(\Delta K=\frac{1}{2}m(v_2^{2}-v_1^{2})=\frac{1}{2}\times5\times(3^{2}-1^{2})=\frac{5}{2}(9 - 1)=\frac{5\times8}{2}=20\space J\). By work - energy theorem \(W_c=\Delta K = 20\space J\)

Step4: Calculate work for option d

Use the formula \(W = Fd\). Here \(F = 5\space N\) and \(d = 5\space m\)
\(W_d=5\times5=25\space J\)

Answer:

D. A net force of \(5\space N\) is applied to an object as it is displaced \(5\space m\)