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Question
four machines in a factory are moving crates around. rank them from greatest (1) to least (4) work done. force: 3f displacement: d time: ½t force: f displacement: d time: t force: ½f displacement: d time: 3t force: ½f displacement: 3d time: t
Step1: Recall Work Formula
Work \( W \) is calculated as \( W = F \times d \) (force times displacement, assuming force and displacement are in the same direction; time does not affect work, only power).
Step2: Calculate Work for Each Machine
- Machine 1: \( F_1 = 3F \), \( d_1 = D \). So \( W_1 = 3F \times D = 3FD \).
- Machine 2: \( F_2 = F \), \( d_2 = D \). So \( W_2 = F \times D = FD \).
- Machine 3: \( F_3 = \frac{1}{2}F \), \( d_3 = D \). So \( W_3 = \frac{1}{2}F \times D = \frac{1}{2}FD \).
- Machine 4: \( F_4 = \frac{1}{2}F \), \( d_4 = 3D \). So \( W_4 = \frac{1}{2}F \times 3D = \frac{3}{2}FD \).
Step3: Rank by Work Done
Compare the work values: \( 3FD > \frac{3}{2}FD > FD > \frac{1}{2}FD \). So the ranking from greatest (1) to least (4) is: Machine 1 (3FD), Machine 4 (\(\frac{3}{2}FD\)), Machine 2 (FD), Machine 3 (\(\frac{1}{2}FD\)).
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- Force: \( 3F \), Displacement: \( D \), Time: \( \frac{1}{2}T \)
- Force: \( \frac{1}{2}F \), Displacement: \( 3D \), Time: \( T \)
- Force: \( F \), Displacement: \( D \), Time: \( T \)
- Force: \( \frac{1}{2}F \), Displacement: \( D \), Time: \( 3T \)