Question

A 1.2 kg block of mass is placed on an inclined plane that has a slope of 30° with respect to the horizontal, and the block of mass is released. A friction force Fk = 2.0 N acts on the block as it slides down the incline. Find the acceleration of the block down the incline.

Answer 1

a≅3.33

Step-by-step explanation:

a=(P×sin∝-Fk)÷1.2

Answer 2

The acceleration of the block down the incline is 3.23 m/s²

Step-by-step explanation:

Fk = frictional force

m = mass

g = acceleration due to gravity

F1 = force in direction of motion

theeta = inclination angle

For Force on inclined plane

F1 = m*g*sin(theeta)

F1 = (1.2)*(9.8)*sin(30)

F1 = 5.88-N

now,

Fnet = F1 - Fk

Fnet = 5.88 - 2

Fnet = 3.88-N

now for acceleration,

Fnet = m*a

a = Fnet / m

a = 3.88 / 1.2

a = 3.23 m/s²