# Homework Help: Distance down an incline

1. Apr 6, 2005

### the_d

A 6 kg box slides down a long, frictionless
incline of angle 34 degrees. It starts from rest at time
t = 0 at the top of the incline at a height 17 m
above the ground. Find the distance the box travels during
the interval 0 < t < 0.6 seconds.

2. Apr 6, 2005

### dextercioby

What are your thoughts...?What kind of movement does the box have?

Daniel.

3. Apr 6, 2005

### the_d

well the distance would be the speed x time but since i am only given time i do not know. the box is moving diogonally, is there some way i can use the height and the angle to find the distance?

4. Apr 6, 2005

### dextercioby

Make a FBD.See that the "driving" force is the tangential component of the gravity force.
U got a linear accelerated motion with a given acceleration & given initial conditions.It's all done.

Daniel.

5. Apr 6, 2005

### the_d

so i could just use X = Xo +Vot + (1/2) at^2?

6. Apr 6, 2005

### marlon

Since dexter gave you so much useful info, let me help you out here. Suppose the incline is downward to the right and the x-axis is along the incline. the y axis is perpendicular to the x-axis. There are two forces acting on the object : gravity and the normal force N.

The clue is to calculate the components along both directions.

$$ma_x = mgsin( 34°)$$
$$ma_y = -mgcos( 34°) + N = 0$$ this 0, because you stay on the incline

You also know the initial position wtr to this frame of reference : 0 in the x-direction and 0 in the y-direction. The initial velocity is also 0.

the formula that you will need is $$x = gsin(34) \frac{t^2}{2}$$
Keep in mind that x is NOT the horizontal distance but the distance travelled along the incline...

regards
marlon

edit : to the OP : lose the m's... my mistake

Last edited: Apr 6, 2005
7. Apr 6, 2005

### the_d

thanx marlon, i had sumtin like that just didnt understand the (t^2/2) part

8. Apr 6, 2005

### dextercioby

It's really sad that a (probably) prolific future PhD stud in physics would make such an erroneous analysis in this simple problem...

Daniel.

9. Apr 6, 2005

### dextercioby

In Marlon's last equation,the mass should have been absent...

And,yes,[itex] \frac{t^{2}}{2} [/tex] times an acceleration is typical for uniformly accelerated motion.U should have known that.

Daniel.

10. Apr 6, 2005

### marlon

WRONG correction dexter, the initial velocity IS ZERO in my chosen reference frame, so is the initial position and so is the force in the y direction

marlon

Last edited: Apr 6, 2005
11. Apr 6, 2005

### the_d

i knew something was wrong because i got an answer like 6 m, and the block couldnt have possibly gone that far in only a fraction of a second. thanx dex