Question

The magnitude of each force is 208 N the force on the right is applied at an angle 36° and the mass of the block is 17 kg the coefficient of friction is 0.149 the acceleration of gravity is 9.8 m/s^2

What is the magnitude of the resulting acceleration?
answer in units m/s^2​

Answer 1

11.06m/s²

Step-by-step explanation:

According to Newtons second law of motion

`\sm F_x = ma_x\\F_m - F_f = ma_x\\mgsin \theta - \mu R mgcos \theta = ma_x\\`

Given

Mass m = 17kg

Fm = 208N

theta = 36 degrees

g = 9.8m/s²

a is the acceleration

Substitute

208 - 0.148(17)(9.8)cos 36 = 17a

208 - 24.6568cos36 = 17a

208 - 19.9478 = 17a

188.05 = 17a

a = 188.05/17

a = 11.06m/s²

Hence the the magnitude of the resulting acceleration is 11.06m/s²