Please wait for the animation to completely load.

A mass is on a frictionless incline as shown in the animation  (position is given in meters and time is given in seconds. You may adjust m, the mass of the block (100 grams < m < 500 grams), and θ, the angle of the incline (10° <  θ < 45° ), and view how these changes affect the motion of the block. Restart.

One of the first things to stress about this type of problem is that, for a suitable set of coordinates, while it is a two-dimensional problem, the motion of the block is one dimensional. Since the motion of the block is down the incline, let's choose that direction for the x axis. Since coordinate axes are perpendicular, let's also call the direction normal to the incline, the y axis. This does two things for us: The net force (and therefore the acceleration) is now on axis (the x axis) and we do not need to decompose the normal force. Check the box and click the "register and play" button to see the free-body diagram for the block and the net force acting on the block.

What force determines the acceleration of the block? It is the part of the gravitational force that is down the incline (mg sin θ). Therefore, the other component of the gravitational force (mg cos θ) must be equal to the normal force since we do not see the block flying off of the incline. The acceleration of the block is g sin θ down the incline.

Now try changing the mass of the block. How do you think the block's acceleration will change as you change the mass?

Now change the angle of the incline. How do you think the angle of the incline affects the acceleration of the block. In the animation you are limited to 10°> θ > 45°. Can you predict, from either the formula or the animation, what will happen to the normal force and the acceleration when θ = 0° and θ = 90°?

Illustration authored by Mario Belloni.
Script authored by Steve Mellema, Chuck Niederriter and modified by Mario Belloni.