An elevator starts from rest with a constant upwards acceleration of 1m in the first 1.4 seconds. A passenger in the elevator is holding a 6.3 kg bundle at the end of the vertical cord. what is the tension in the cord as the elevator accelerates. the acceleration of gravity is 9.8 m/s squared
To find the tension in the cord, we can use Newton's second law of motion, which states that the net force on an object is equal to the mass of the object multiplied by its acceleration.
In this case, the object is the bundle held by the passenger, and the net force acting on it is the tension in the cord. The acceleration of the bundle is the same as the elevator's acceleration, given as 1 m/s^2.
Let's break down the problem step by step:
Step 1: Find the gravitational force acting on the bundle.
The gravitational force (Fg) can be calculated using the formula: Fg = m * g, where m is the mass of the object and g is the acceleration due to gravity (9.8 m/s^2). Given that the mass of the bundle is 6.3 kg, we can calculate Fg.
Fg = 6.3 kg * 9.8 m/s^2 = 61.74 N
Step 2: Find the net force on the bundle.
The net force (Fnet) acting on the bundle is the difference between the tension in the cord and the gravitational force.
Fnet = Ftension - Fg
Step 3: Use Newton's second law to relate the net force to the acceleration.
According to Newton's second law, Fnet = m * a, where m is the mass of the object and a is the acceleration. Rearranging the equation, we get Ftension = m * a + Fg.
Step 4: Calculate the tension in the cord.
Ftension = (6.3 kg * 1 m/s^2) + 61.74 N
Ftension = 6.3 N + 61.74 N
Ftension = 68.04 N
Therefore, the tension in the cord as the elevator accelerates is 68.04 N.