A ball of mass M is suspended by a thin string (of negligible mass) from the ceiling of an elevator. uploaded image

The vertical motion of the elevator as it travels up and down is described in the statements below. Indicate for each of the situations described the relation between value of the tension in the cable, T, and the weight of the ball, Mg, or whether one Cannot tell. (Assume that there is no air, i. e., neglect the buoyancy effect of the air.)

Answer Choices: T>Mg, T
(a) The elevator is traveling upward and its upward velocity is decreasing as it nears a stop at a higher floor.
(b) The elevator is traveling upward and its upward velocity is increasing as it begins its journey towards a higher floor.
(c) The elevator is traveling downward and its downward velocity is decreasing as it nears a stop at a lower floor.
(d) The elevator is traveling downward at a constant velocity
(e) The elevator is traveling downward and its downnward velocity is increasing
(f) The elevator is stationary and remains at rest.

(a) The elevator is traveling upward and its upward velocity is decreasing as it nears a stop at a higher floor.  T > mg

(b) The elevator is traveling upward and its upward velocity is increasing as it begins its journey towards a higher floor. T > mg

(c) The elevator is traveling downward and its downward velocity is decreasing as it nears a stop at a lower floor. T < mg

(d) The elevator is traveling downward at a constant velocity. T = mg

(e) The elevator is traveling downward and its downward velocity is increasing. T < mg

(f) The elevator is stationary and remains at rest. T = mg

Explanation:

To answer this question, consider all the forces acting on the elevator.

The mass of the ball acting downwards due to gravity = mg

The tension on the string depends on upward or downwards force on the ball. T = m(a+g)

where a is acceleration and increase in velocity causes increase in acceleration, and vice versa. (a = v/t)

(a) The elevator is traveling upward and its upward velocity is decreasing as it nears a stop at a higher floor.

If the upward velocity is decreasing, its acceleration is also decreasing, and acceleration is not equal to Zero

T = m(a+g) > mg

(b) The elevator is traveling upward and its upward velocity is increasing as it begins its journey towards a higher floor.

If the upward velocity is increasing, its acceleration is also increasing.

Then, T = m(a+g) > mg

(c) The elevator is traveling downward and its downward velocity is decreasing as it nears a stop at a lower floor.

If the downward velocity is decreasing, its acceleration is also decreasing, and acceleration is not equal to Zero

T = m(a-g) < mg

(d) The elevator is traveling downward at a constant velocity

At constant velocity, acceleration is zero, because acceleration is the rate of change of velocity.

T = m(0+g) = mg

(e) The elevator is traveling downward and its downward velocity is increasing

If the downward velocity is increasing, its acceleration is also increasing

T = m(a-g) < mg

(f) The elevator is stationary and remains at rest.

if the elevator is at rest, its acceleration is zero

T = m(0+g) = mg