Question

Consider a 71.0−kg man standing on a spring scale in an elevator. Starting from rest, the elevator ascends, attaining its maximum speed of 1.76 m/s in 0.750 s. It travels with this constant speed for the next 5.00 s. The elevator then undergoes a uniform acceleration in the negative y direction for 1.60 s and comes to rest. What does the spring scale register before the elevator starts to move?

Answer 1

Before the elevator starts moving, the spring scale would register the weight of the 71.0-kg man, which is 695.8 N, as the elevator is not accelerating at this point.

Step-by-step explanation:

When the 71.0-kg man is standing on a spring scale in an elevator before the elevator starts to move, the scale will register his weight. This is because the elevator is not accelerating, so the only force acting on the man is gravity, which is pulling him downward. The scale provides an equal and opposite force upward, which is known as the normal force. The reading on the scale is the normal force exerted by the scale, which is equal to the man's weight (the force of gravity acting on his mass). The weight can be calculated using the formula W = mg, where m is the man's mass and g is the acceleration due to gravity (9.8 m/s² on Earth).

To find the weight of the man before the elevator starts moving:

W = mg = 71.0 kg × 9.8 m/s² = 695.8 N.

Therefore, the scale would register 695.8 N when the elevator is stationary.