A 75-kg person is standing on a scale in an elevator. What is the reading of the scale in newtons if the elevator is (a) at rest, and (b) moving up with a constant velocity of 2.0 m/s?
The correct answer and explanation is :
Solution:
The scale measures the normal force (N) exerted on the person, which is equal to their apparent weight. The weight of the person is given by:
[
W = mg
]
where:
- ( m = 75 ) kg (mass of the person)
- ( g = 9.81 ) m/s² (acceleration due to gravity)
[
W = 75 \times 9.81 = 735.75 \text{ N}
]
(a) Elevator at Rest
When the elevator is at rest, there is no additional acceleration acting on the person. The only force acting is gravity, balanced by the normal force:
[
N = mg = 735.75 \text{ N}
]
So, the scale reads 735.75 N.
(b) Elevator Moving Up at Constant Velocity (2.0 m/s)
When an object moves with constant velocity, its acceleration is zero. According to Newton’s Second Law:
[
N = mg + ma
]
Since ( a = 0 ), the equation simplifies to:
[
N = mg = 735.75 \text{ N}
]
So, the scale still reads 735.75 N.
Explanation (300 Words)
The key concept in this problem is Newton’s Second Law, which states that the net force acting on an object is given by:
[
F_{\text{net}} = ma
]
where ( a ) is the acceleration. A scale measures the normal force, which changes if there is additional acceleration.
Case (a): Elevator at Rest
When the elevator is not moving, the only forces acting are:
- Gravity (( mg )), pulling the person downward.
- The normal force (( N )), exerted by the scale, pushing upward.
Since the person is not accelerating, these forces balance each other, making the scale read 735.75 N.
Case (b): Moving Up at Constant Velocity
When moving at constant velocity, the acceleration is zero. According to Newton’s First Law, an object in motion stays in motion unless acted upon by an external force. Since there is no acceleration, the forces remain unchanged, meaning the normal force (( N )) is still equal to ( mg ), and the scale reads 735.75 N.
Thus, whether at rest or moving at constant velocity, the scale reading remains 735.75 N.