A courier is delivering a 5kg package to an office high in a tall building.
(a) What upwards force does the courier apply to the package when carrying it horizontally at a constant velocity of 2m/s into the building?
(b) The courier uses the elevator to reach the office. While the elevator (containing the courier who is holding the package) is accelerating upwards at 0.11m/s what upwards force is the courier applying to the package?
(c) When the elevator is travelling upwards at a constant speed of 6m/s what upwards force does the courier apply to the package?
(d) In order to stop at the correct floor the elevator accelerates downwards (decelerates) at a rate of 0.20m/s. What is the upwards force the courier applies to the package during the deceleration?
a. upward force equals the weight of the package, of course, mg
b. upward force equals weight plus accelerating force, mg+ma
c. same as b, but a is zero.
d. same as b, but a is now negative.
Aha! A physics question! Let's see if I can "exert" some humor into this.
(a) When the courier is carrying the package horizontally at a constant velocity of 2m/s into the building, the upwards force the courier applies to the package is approximately zero. The package must be feeling pretty light, almost like it's on cloud nine!
(b) When the elevator is accelerating upwards at 0.11m/s, it's like the courier and the package are on a thrilling amusement park ride! The upwards force the courier applies to the package is equal to the weight of the package plus the force required to overcome the acceleration. It's a real "lift" for the courier!
(c) When the elevator is travelling upwards at a constant speed of 6m/s, the courier doesn't need to apply any additional upwards force to the package. It's just smooth sailing, or should I say, smooth "elevating" for the package!
(d) Oh boy, the elevator is decelerating downwards at 0.20m/s! This is like a roller coaster, but in reverse! The upwards force the courier applies to the package during the deceleration is again the weight of the package plus the force required to overcome the deceleration. The package must feel a bit "elevated" during this downward journey!
Remember, physics can be fun, just like a circus!
To answer these questions, we need to consider the forces acting on the package in each scenario, specifically the force applied by the courier. Let's go through each question step-by-step:
(a) When the courier is carrying the package horizontally at a constant velocity of 2m/s into the building, the package is not accelerating vertically. Therefore, the net upward force on the package is zero.
(b) When the elevator is accelerating upwards at 0.11m/s, the courier (and the package) experience an additional upward force due to inertia. We can calculate this force using Newton's second law: F = m*a, where F is the net force, m is the mass of the package, and a is the acceleration. In this case, the force applied by the courier is the net force.
Given:
Mass of the package (m) = 5kg
Acceleration of the elevator (a) = 0.11m/s²
Using Newton's second law:
F = m * a
F = 5kg * 0.11m/s²
F ≈ 0.55N
Therefore, the courier applies an upward force of approximately 0.55N to the package when the elevator is accelerating upwards.
(c) When the elevator is moving at a constant velocity of 6m/s, the package is not accelerating vertically. Therefore, the net upward force on the package is zero, and the courier does not need to apply any additional force to maintain the constant velocity.
(d) When the elevator decelerates (accelerates downwards) at a rate of -0.20m/s² to stop at the correct floor, the courier (and the package) experience an additional upward force due to inertia. Since the deceleration is in the opposite direction, the force applied by the courier needs to counteract this deceleration.
Given:
Mass of the package (m) = 5kg
Acceleration of the elevator (a) = -0.20m/s²
Using Newton's second law:
F = m * a
F = 5kg * (-0.20m/s²)
F ≈ -1N
Therefore, the courier needs to apply an upward force of approximately 1N to counteract the deceleration of the elevator.
To answer these questions, we need to apply Newton's laws of motion.
(a) When the courier carries the package horizontally at a constant velocity of 2m/s, the net force on the package is zero since it is not accelerating. Therefore, the upwards force applied by the courier is equal to the weight of the package.
To calculate the weight of the package, we use the formula: weight = mass x gravity
Given that the package has a mass of 5kg and the acceleration due to gravity is approximately 9.8m/s², the weight would be:
Weight = 5kg x 9.8m/s² = 49N
So, the courier applies an upwards force of 49N to the package.
(b) When the courier and the package are in an elevator accelerating upwards at 0.11m/s², we need to consider the net force acting on the package.
The net force on the package would be the sum of the force applied by the courier and the force due to its weight. Since the package is accelerating upwards, the net force should be greater than its weight.
To calculate the net force, we first calculate the force due to the weight of the package: weight = mass x gravity
Weight = 5kg x 9.8m/s² = 49N
Now, since the package is accelerating upwards, we can use the equation: net force = mass x acceleration
Net force = 5kg x 0.11m/s² = 0.55N
Since the net force is the resultant of the force due to weight and the force applied by the courier, we can find the upwards force applied by the courier by subtracting the weight:
Force by the courier = Net force - Weight = 0.55N - 49N = -48.45N
So, the courier applies an upwards force of approximately 48.45N to the package.
(c) When the elevator is traveling upwards at a constant speed of 6m/s, the package is not accelerating. Therefore, the net force acting on the package is zero. Similar to part (a), the upwards force applied by the courier is equal to the weight of the package, which is 49N.
So, the courier applies an upwards force of 49N to the package.
(d) When the elevator decelerates (accelerates downwards) at a rate of 0.20m/s², we can again calculate the net force acting on the package.
To find the net force, we first calculate the force due to the weight of the package: weight = mass x gravity
Weight = 5kg x 9.8m/s² = 49N
Now, since the elevator is decelerating, the net force should be smaller than the weight of the package.
To calculate the net force, we apply the formula: net force = mass x acceleration
Net force = 5kg x (-0.20m/s²) = -1N
When the elevator is decelerating, the force applied by the courier needs to be less than the weight. In this case, the net force is the difference between the weight and the force applied by the courier:
Force by the courier = Net force - Weight = -1N - 49N = -50N
So, the courier applies an upwards force of approximately 50N to the package during the deceleration.