Question

The inclined plane and horizontal shown above both have a coefficient of kinetic friction of 0.10. 

 a) If the 8.00 kg mass is suddenly released from rest, find how far the system moves in 3.0 seconds.

 b) What is the tension in the cable while the masses are sliding?

Answer 1

m1 = 8 kg ,

m2 = 14 kg

theta = 50 degree

uk = 0.1

a)

the accelration , a = net force /effective mass

a = ( m2 * g * sin(theta) - uk * ( m2 * g * cos(theta) + m1 * g))/(m1 + m2)

a = ( 14 * 9.81 * sin(50) - 0.1 * ( 14 * 9.81 * cos(50) + 8 * 9.81)) /(8 + 14)

a = 4.02 m/s^2

time taken , t = 3 s

s = 0 + 0.5 * a * t^2

s = 0 + 0.5 * 4.02^2 * 3^2 m

s = 72.7 m

b)

the tension in the cable , T = uk * m1 * g + m1 * a

T = 0.1 * 8 * 9.81 + 8 * 4.02 N

T = 40 N