The tension on a string from which a 5.00kg object is suspended in an elevator is T=(44.0j)N. What is the acceleration of the elevator? Specify the acceleration in unit vector notation.
T - Mg = M a with positive force upward.
a = T/m - g = 8.8 - 9.8 = -1.0 m/s^2
Multiply that by a j vector to denote the downward acceleration
To find the acceleration of the elevator, we can use Newton's second law of motion. The force equation for this scenario is given as T - Mg = Ma, where T is the tension on the string, M is the mass of the object, and a is the acceleration of the elevator.
Given T = (44.0j)N (in unit vector notation) and M = 5.00kg, we can substitute these values into the force equation: (44.0j)N - (5.00kg)(9.8m/s^2) = (5.00kg)a.
Simplifying this equation, we get:
44.0j - 49.0j = 5.00kga
Calculating the vector subtraction, we find that the tension force and the weight force have opposite directions, resulting in a net force in the negative y-direction:
-5.00jN = 5.00kga
To isolate a, we divide both sides of the equation by 5.00kg:
a = (-5.00jN) / 5.00kg
a = -1.0j m/s^2
Therefore, the acceleration of the elevator is -1.0 m/s^2, in the downward direction (denoted by j) in unit vector notation.