A weightlifter holds a barbell stationary above their head. They then lower the barbell slowly back down to the ground at a constant velocity. Explain, using Newton's laws, why the barbell moves at a constant velocity as it is lowered, and describe what happens to the forces acting on the barbell when it is first held stationary above the weightlifter's head.
Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).
A weightlifter lifts a heavy barbell from the ground and holds it stationary above their head. They then lower it slowly back to the ground at a constant velocity.
Model answer (5 marks)
When the barbell is held stationary the only forces are its weight,
(W=mg) downward, and the upward force from the weightlifter’s hands, (F_{ ext{up}}). For the barbell to remain at rest the resultant force must be zero, so (F_{ ext{up}}=W). This is a direct application of Newton’s First Law – an object at rest stays at rest when the net force is zero.
When the barbell is lowered at a constant velocity the situation is the same: the barbell’s weight still pulls it downwards, and the weightlifter’s hands still push upward. Because the velocity is constant, the acceleration is zero, so by Newton’s Second Law the net force must again be zero. Therefore (F_{ ext{up}}) during the lowering is equal to the weight of the barbell.
At the moment the barbell is first held above the head the upward force from the hands is exactly equal to the weight; when it is lowered the upward force remains equal to the weight, keeping the net force zero and maintaining constant velocity.
(W=mg) downward, and the upward force from the weightlifter’s hands, (F_{ ext{up}}). For the barbell to remain at rest the resultant force must be zero, so (F_{ ext{up}}=W). This is a direct application of Newton’s First Law – an object at rest stays at rest when the net force is zero.
When the barbell is lowered at a constant velocity the situation is the same: the barbell’s weight still pulls it downwards, and the weightlifter’s hands still push upward. Because the velocity is constant, the acceleration is zero, so by Newton’s Second Law the net force must again be zero. Therefore (F_{ ext{up}}) during the lowering is equal to the weight of the barbell.
At the moment the barbell is first held above the head the upward force from the hands is exactly equal to the weight; when it is lowered the upward force remains equal to the weight, keeping the net force zero and maintaining constant velocity.
Examiner tips
- Show the forces explicitly (weight and upward force) and state they are equal for a stationary barbell.
- Explain that constant velocity means zero acceleration, so net force zero – use Newton’s Second Law or First Law as appropriate.
- Mention that the upward force during lowering is still equal to the weight, not greater or lesser.
- Keep the answer concise – 5 marks, so one sentence per point is enough.
Common mistakes
- Writing that the upward force is greater than the weight when the barbell is lowered – this would imply acceleration.
- Forgetting to mention that the net force is zero during constant velocity.
- Using the wrong law (e.g., citing Newton’s Third Law instead of First/Second).
Mark scheme (5 marks)
- When the barbell is held stationary, the forces acting on it are balanced (resultant force is zero).
- The weight of the barbell acts downward and the upward force from the weightlifter's hands is equal in magnitude.
- Newton's First Law states that an object remains stationary or moves at constant velocity when the resultant force acting on it is zero.
- When the barbell is lowered at constant velocity, the forces on it are balanced / the resultant force is zero.
- The upward force from the weightlifter's hands is less than the weight when stationary is compared to the lowering phase, OR correctly states that during lowering the weightlifter's hands exert an upward force equal to the weight of the barbell.
Key terms in this question
Related
- All Eduqas A-Level Physics revision notes →
- How to answer a "Explain" question →
- Decode the mark scheme abbreviations →
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