## SAT Physics Subject Test

## Chapter 3 Newton”s Laws

### INCLINED PLANES

An **inclined plane** is basically a ramp. If an object of mass, *m,* is on the ramp, then the force of gravity on the object, **F**_{w} = *m***g**, has two components: one that”s parallel to the ramp (*mg* sin*θ*) and one that”s normal to the ramp (*mg* cos*θ*), where *θ* is the incline angle. The force driving the block down the inclined plane is the component of the block”s weight that”s parallel to the ramp: *mg* sin*θ*.

21. A block slides down a frictionless, inclined plane that makes a 30˚ angle with the horizontal. Find the acceleration of this block.

**Which Component?**

To avoid drawing a free

body diagram for every

inclined plane problem,

remember that *mg* sin** θ** is

the component of

gravity down the inclined

plane by thinking “sine”

equals “sliding.”

Here”s How to Crack It

Let *m* be the mass of the block, so the force that pulls the block down the incline is *mg* sin*θ*, and the block”s acceleration down the plane is

22. A block slides down an inclined plane that makes an angle *θ* with the horizontal. If the coefficient of kinetic friction is µ, find the acceleration of the block.

Here”s How to Crack It

First draw a free-body diagram. Notice that in the diagram shown below, the weight of the block, **F**_{w} = *m***g**, has been written in terms of its scalar components: **F**_{w} sin *θ* parallel to the ramp, and **F**_{w} cos *θ* normal to the ramp.

The force of friction, **F**_{f}, that acts up the ramp (opposite to the direction in which the block slides) has magnitude *F*_{f} = µ*F*_{N}. But the diagram shows that *F*_{N} = *F*_{w} cos*θ*, so *F*_{f} = µ(*mg* cos*θ*). Therefore, the net force down the ramp is

F_{w} sin*θ* – F_{f} = *mg* sin*θ* – µ*mg* cos*θ* = *mg*(sin *θ*– µcos *θ*).

Then, setting *F*_{net} equal to *ma*, we solve for *a*: