Acceleration is 7 ms2 I just want to know how to get these a

16 kg Examples A 5 kg bilock slides Example 5 A 5 kg block slides down an inclined surface (36.87°) that has ,-0.25 . A cord. attached to this block, passes over a flywheel (M=16 kg, R = 0.2 m, and radius of gyration k-0.1 m) and is attached to a 1.0 kg block, as shown. Find a) acc. of each block, and b) the tension in each cord. As sun c i 53 =.25 Dish

Solution

Here ,

for the acceleration of the system is a

moment of inertia of pulley is I = 0.5 * M * R^2

Using second law of motion

a = net force/effective mass

a = (5* sin(36.87 degree) - 1 - 0.25 * 5 * cos(36.87 degree) ) * 9.8/(5 + 1 + 0.5 * 16 * 0.2^2/0.2^2 )

a = 0.70 m/s^2

the acceleration of the blocks is 0.70 m/s^2

b)

for the 1 kg block

T - m * g = m * a

T = 1 * (0.7 + 9.8)

T = 10.5 N

the tension in the string is 10.5 N