How far does the 6 kg box slide down the incline in 0.6 seconds?

[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.

 

[dextercioby]

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

Daniel.

 

[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?

 

[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.

 

[the_d]

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

 

[marlon]

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.

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 traveled along the incline...

regards
marlon

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

 

[the_d]

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

 

[dextercioby]

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.

$$a_x = mgsin( 34°)$$
$$a_y = -mgcos( 34°) + N = 0$$ 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 $$x = mgsin(34) \frac{t^2}{2}$$
Keep in mind that x is NOT the horizontal distance but the distance traveled along the incline...

regards
marlon

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

Daniel.

 

[dextercioby]

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

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.

 

[marlon]

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

Daniel.

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

 

[the_d]

i knew something was wrong because i got an answer like 6 m, and the block couldn't have possibly gone that far in only a fraction of a second. thanks dex