A weightlifter lifts a 275 kg barbell from the ground to a height of 2.4 m. How much work has the weightlifter done in lifting the barbel and how much work is required to hold a weight at that height?
Correct answer: C. 6468J and 0J respectively
- A. 3234J and 0J respectively
- B. 3234J and 3234J respectively
- C. 6468J and 0J respectively
- D. 6468J and 6468J respectively
Explanation
To calculate the work done in lifting the barbell, use the formula Work = Force x Distance. The force required is equal to the weight of the barbell, which is mass times the acceleration due to gravity (9.8 m/s²). Thus, Force = 275 kg x 9.8 m/s² = 2695 N. The work done is then Work = 2695 N x 2.4 m = 6468 J. Once the barbell is lifted, holding it at that height requires no additional work because there is no displacement. Therefore, the correct answer is 6468J and 0J respectively. Options A and B miscalculate the work done to lift the barbell, while Option D incorrectly states that work is required to hold the barbell at a height.
Last updated
About Force and Motion
Displacement, velocity and acceleration describe motion, including the horizontal and vertical components of projectile motion. The chapter connects these quantities with Newton's laws, momentum and impulse, then applies conservation of momentum to collisions, distinguishing elastic collisions from inelastic ones.
Practise Force and Motion
1,416 free Force and Motion MCQs from Physics, each with the correct answer and an explanation. Unlimited attempts, no account needed.
Exams that ask Physics questions like this
Physics is on 13 papers prepared for on TestUstad, and all of them draw the same bank, so this question is worth knowing for every one of them.
Related questions
(1, 1, 1) . (2, 2, 2)
1st law of motion gives the definition of:
2nd law of motion measures which quantity?
7.0 kg bowling ball experience a net force of 5.0N. What will be its acceleration?
A 0.2 kg rubber ball is dropped from the window of a building. It strikes the sidewalk below at 30m/s and rebounds up at 20m/s. The impulse on the ball during the collision is: