Name: Answer Key Honors Physics 5th 6th

5.4 Finding acceleration of an object on an incline plane

A block is released from rest and slides down a frictionless, incline plane angled at 40° that is 20m long. Find:

a) The acceleration on the block

b) The time it takes to move

Follow these steps to solve problems like this.

Step 1: Free-body diagram:

Step 2: Break vectors into components. Because we have an incline, we use

inclined axes, one parallel and one perpendicular to the incline:

Step 3: Write Newton's second law for each axis. The acceleration is entirely directed parallel to the plane, so

perpendicular acceleration can be written as zero:

mg sin θ – 0 = ma.

FN – mg cos θ = 0

Another way to look at this is, the downslope force is equal to:

Fdown= Fnsinθ = (mg)sinθ and Fdown = ma

So we can say that: ma=(mg)sinθ

Solving for acceleration: please note correction to this formula!!

( )

Step 4: Solve algebraically for a. This can be done without reference to the second equation. (In problems with

friction, use Ff = μFN to relate the two equations.) NOTE: The mass of the object cancels out. So the mass of the

object will have no effect on the acceleration of the object.

a = g sin θ a = 9.8 sin 40°

The acceleration of the object is: __________6.3m/s2______________________________

Step 5: To find the time, use K2: ⁄

vo = 0 a= (your answer from part 4) 20m= 0t + ½ 6.3 m/s2t2

d= 20m t= ? √

The time it takes the object to move down the incline is: t = 2.0s

Practice Problems:

For each of the following problems:

a) The acceleration on the block

b) The time it takes to move

Assume that the plane is frictionless unless stated, and the block is starting from rest.

1.

First calculate the angle using trig,

.

Then calculate the downslope acceleration ⁄

Then use K2 to solve for time: ( )

√ ⁄

2.

In this next problem, there is a friction force involved. So the NET FORCE has to be found down the

incline. The net force is the downslope force minus the friction force. Remember that W=mg

First, calculate the downslope force:

( )

Next, find the Net Force

Now that we know the Net Force, we can solve for acceleration using N2, Fnet=ma

Solving for acceleration - ⁄

Then use K2 to solve for time: ( )

√ ⁄

3.

This problem is very similar to problem 2, with the exception that the friction force is not given. So first

calculate the friction force. We know that the Ffriction=μFn, so first calculate the normal force.

Fn= Wcosθ = 98 · cos 40° = 75N

Calculate the friction force:

Ffriction=0.25·75N = 18.8N

Then solve for the downslope force

( )

Next, find the Net Force

Now that we know the Net Force, we can solve for acceleration using N2, Fnet=ma

Solving for acceleration - ⁄

Then use K2 to solve for time: ( )

√ ⁄

4.

First calculate the angle using trig,

.

We also need to find the downslope distance. This can be found using the Pythagorean theorem:

√

Since there is no friction, the downslope acceleration can be calculated as:

⁄

Then use K2 to solve for time: ( )

√ ⁄

5.

Since there is no friction, the downslope acceleration can be calculated as:

⁄

down the plane

Since there is an initial velocity, it needs to be taken into account. We will still use K2: ⁄

( )

To solve this, we need to dust off the old quadratic equation!

√

Rewrite the problem in quadratic form: 3t2

+ 5t -20 = 0

√ ( )

Since only 1 result yields a positive number (and we haven’t invented time travel) t = 1.9s

6.

Since there is no friction, the downslope acceleration can be calculated as:

⁄

down the plane

This time, the initial velocity is up the plane, making it negative. We will still use K2: ⁄

( )

To solve this, we need to dust off the old quadratic equation!

√

Rewrite the problem in quadratic form: 3t2

- 5t - 20 = 0

( ) √ ( )

Since only 1 result yields a positive number (and we haven’t invented time travel) t = 3.5s

7.

If you want to skip this problem, don’t worry. It won’t count against you. I’m not posting all the steps.

The time is solved using the quadratic formula we used prior. There WILL NOT be any problems like

this on the test!

This one is complicated. Since the direction of the friction force changes depending on whether the block

is sliding up or down the plane, the block's acceleration is NOT constant throughout the whole problem.

So, unlike problem #7, this one can't be solved in a single step. Instead, in order to use kinematics

equations, you must break this problem up into two parts: up the plane and down the plane. During each

of these individual parts, the acceleration is constant, so the kinematics equations are valid.

o up the plane:

a = 6.8 m/s2, down the plane.

t = 0.4 s before the block turns around to come down the plane.

o down the plane:

a = 1.5 m/s2, down the plane.

t = 5.2 s to reach bottom.

So, a total of t = 5.6 s for the block to go up and back down.