If the motor draws in the cable at a rate of v = (0.05 t 2 ) m / s , where t is
ID: 2994022 • Letter: I
Question
If the motor draws in the cable at a rate of v = (0.05 t2) m/s, where t is in seconds, determine the tension developed in the cable when t = 5s . The crate has a mass of 22kg and the coefficient of kinetic friction between the crate and the ground is ?k = 0.23.
Explanation / Answer
If we differentiate the fiven expression for velocity with respect to time we will get the expression for acceleration.
So, we get a = (0.05)*(2t) = 0.1t m/sec^2
At t=5 sec a = 0.1*5 = 0.5 m/sec^2
So, net force applied = mass*acceleration = 22*0.5 = 11 Newtons
Frictional force on the block = mu*N = 0.23*22*9.81 = 49.63 N
Net force = Applied force - Frictional force
Hence,
Applied force = 11+46.93 = 60.63 N
This applied force is the value of tension