Question

correct, explain why. b4 - rt08: equal forces on boxes - work done on box in the figures below, identical boxes of mass 10 kg are moving at the same initial velocity to the right on a flat surface. the same magnitude force, f, is applied to each box for the distance, d, indicated in the figures rank the work done on the box by f while the box moves the indicated distance. explain your reasoning.

Explanation:

Step1: Recall the work formula

The work formula is \(W = Fd\cos\theta\), where \(F\) is the force, \(d\) is the displacement, and \(\theta\) is the angle between the force and the displacement.

Step2: Analyze each case

• Case A: \(\theta = 0^{\circ}\), \(W_A=F\times5\times\cos0^{\circ}=5F\)
• Case B: \(\theta = 0^{\circ}\), \(W_B=F\times10\times\cos0^{\circ}=10F\)
• Case C: \(\theta\) is acute, \(\cos\theta< 1\), \(W_C=F\times5\times\cos\theta<5F\)
• Case D: \(\theta = 90^{\circ}\), \(\cos\theta = 0\), \(W_D=F\times10\times\cos90^{\circ}=0\)
• Case E: \(\theta = 180^{\circ}\), \(\cos\theta=- 1\), \(W_E=F\times5\times\cos180^{\circ}=-5F\) (magnitude is \(5F\))
• Case F: \(\theta = 90^{\circ}\), \(\cos\theta = 0\), \(W_F=F\times5\times\cos90^{\circ}=0\)

Step3: Rank the work

The magnitudes of the work: \(W_B>W_A = |W_E|>W_C>W_D = W_F\)

Answer:

\(B>A = E>C>D = F\)