A 7.0-N force parallel to an incline is applied to a 1.0-kg crate. The ramp is tilted at 20° and is frictionless.

(a) What is the acceleration of the crate?

(b) If all other conditions are the same but the ramp has a friction force of 1.9 N, what is the acceleration?

a) 0.1m/s^2 b) 0.12m/s^2

Explanation:

a) Using the concept of friction to solve the question,

First we need to know all the forces acting on the body on the inclined plane

Forces acting parallel to the plane are the moving force (Fm), frictional force (Ff) and mgsintheta (resolving weight along the plane).

Force acting perpendicular to the plane are weight (mgcostheta) and the normal reaction.

Using the formula

Summation of forces along the plane = mass * acceleration

Fm+mgsintheta - Ff = ma ... (1)

Fm = 7.0N

Mass = 1.0kg

Ff=0 (since the movement is frictionless)

Substituting the datas in (1)

7.0 + 1*10sin20-0 = 1.0a

7.0+3.42=1.0a

a = 1.0/10.42

a = 0.1m/s^2

b) Now that the fictional force is 1.9N, the acceleration will change. Using the same formula in (a)

7.0+1*10sin20-1.9 = 1.0a

7.0+3.42-1.9=1.0a

8.52 = 1.0a

a = 1.0/8.52

a=0.12m/d^2