Question

A 275N crate is sliding down a 35 degrees incline. If the force of friction along the incline is 96N, what is the acceleration of the crate?

Answer 1

The projection of the gravity force on the crate along the incline (the force that pushes the crate down) is:

$$F_{down} = 275\cdot \sin 35\degree$$

The force that pushes it up along the incline is the friction force:

$$F_{up} = 96N$$

Thus, according to the second Newton's law, their difference is:

$$F_{down} - F_{up} = ma\\ a = \dfrac{F_{down} - F_{up} }{m}$$

where $m$ is the mass of the crate, and $a$ is its acceleration.

The mass can be found from the weight:

$$m = \dfrac{275N}{9.8N/kg}$$

where $9.8N/kg$ is the gravitational acceleration.

Thus, obtain:

$$a = \dfrac{ 275\cdot \sin 35\degree}{\dfrac{275N}{9.8N/kg}} - \dfrac{96N}{\dfrac{275N}{9.8N/kg}} \approx 2.2m/s^2$$

Answer. 2.2 m/s^2.