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Newton's Laws of Motion Study Guide (page 4)

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Updated on Sep 28, 2011

Example

Consider an object accelerating and decelerating due to gravity on an inclined plane that has a kinetic frictional coefficient μk. Find the expression of the acceleration along the plane in the two cases mentioned.

Solution

Start with the free-body diagram for the first case, applying the conditions of equilibrium appropriate to each direction and then calculate the acceleration (see Figure 5.12).

Then find out the components of the weight along the plane and perpendicular to the plane: Wx and Wy.

Wx = W sin α = m g sin α

Wx = W cos α = m g cos α

Then write Newton's second law for directions along the plane and perpendicular to the plane. Again, we will work with each direction so we can use a scalar equation.

y-axis

ay = 0

m ay = NWy

0 = NWy

N = Wy = m g cos α

x-axis

m ax = Wxfk = Wx – μk N

m ax = m g sin α – μk m g cos α

The acceleration is:

axDOWN = g(sin α - μkcos α)

In the second case, when the object is moving up the inclined plane, we have a different free-body diagram. In order for the object to move up, imagine you gave it an intial push. All forces are the same except the fricitonal force, which acts against motion. Then apply Newton's second law along the plane and perpendicular to the plane as follows:

y-axis

ay= 0

m ay = NWy

0 = NWy

N = Wy = m g cos α

x-axis

m · ax = Wx + fk = Wx + μkN

m · ax = m g sin α + μkm g cos α

And the acceleration is:

ax UP = – g(sin α + μk cos α)

Practice problems of this concept can be found at: Newton's Laws of Motion Practice Questions

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