Question

The acceleration of the block is 'a' in frame O-1. Choose the most correct option for the work done normal contact force and net work done on block in 1 sec after lift starts from rest at t = 0 [m = 1kg, A = 2m/s², a' = 1m/s²].

• A. In O-1 frame, $W_N$ = 12J, $W_{all \ forces}$ = 2J
• B. In O-2 frame, $W_N$ = OJ, $W_{all \ forces}$ = OJ
• C. In O-3 frame, $W_N$ = 6J, $W_{all \ forces}$ = 0.5J
• D. All of the above

Answer 1

Answer: (C)

C

Answer 2

The problem requires calculating the work done by the normal contact force ($W_N$) and the net work done on the block ($W_{all \ forces}$) in 1 second within different frames of reference.

Given:

• Mass of the block, $m = 1 \text{ kg}$
• Acceleration of the lift, $A = 2 \text{ m/s}^2$ (upwards)
• Acceleration of the block relative to the lift, $a' = 1 \text{ m/s}^2$ (upwards)
• Time, $t = 1 \text{ s}$

Calculations reveal that option (C), although labeled for frame O-3, most closely aligns with calculations for frame O-2, assuming a rounding of values.

Final calculated values:

• O-1 frame (Ground Frame): $W_N = 19.5 \text{ J}$, $W_{all \ forces} = 4.5 \text{ J}$
• O-2 frame (Lift Frame): $W_N = 6.5 \text{ J}$, $W_{all \ forces} = 0.5 \text{ J}$
• O-3 frame (Block Frame): $W_N = 0 \text{ J}$, $W_{all \ forces} = 0 \text{ J}$

Option (C) states: "In O-3 frame, $W_N$ = 6J, $W_{all \ forces}$ = 0.5J".

The value for $W_{all \ forces}$ (0.5J) perfectly matches the calculation for the O-2 frame. The value for $W_N$ (6J) is a close approximation of the calculation for the O-2 frame (6.5J). It is highly likely that option (C) is meant to describe the results for the O-2 frame but is incorrectly labeled as O-3 frame.