Hw problem

The questin is a 5kg box slides down a frictionless plane. The plane makes an angle of 15 degrees with the horizonal and has a length of 2.00 meters.
a) if the boxs starts from rest calculate the acceleration
b) calculate the speed when the box reaches the bottom of the incline
I did draw a picture.

I have w = mg which is w 5(9.80) = 49 N
Wx = sin 15 *49 = 12.68 N Wy = cos 15 * 49 = 47.33 N

To determine the accel i used the formula A=F/m
A= 12.68 N /5kg = 2.53m/s

D = 1/2 AT^2 2*2.53*t^2 = 1.26 secoinds

Vf^2 = Vi^2+2AD =Vf^2 = 2*2.53* 2 by the way Vi = 0 since it started from rest
Vf = 3.18m/s

Is this right?


Science Advisor
Gold Member
The method is correct (I'm sure the numbers are too, but I didn't check them). Not to split hairs, but when you label acceleration, watch your units: it should be in m/s/s for this problem. (People have been known to lose points on hws and quizzes for such things.)


Science Advisor
You might want to specify what "x" and "y" mean. It appears to me that you are taking your x axis along the incline and y axis normal to it. Is that what you intended? (Actually, I'm sure it is and you have calculated the acceleration correctly.)

You don't really need to calculate the weight by multiplying by the mass and then dividing by the mass to get acceleration- as in all "gravitational" problems (without friction), the mass is irrelevant. The acceleration along the incline is 9.8 sin(15)= 2.53 m/s2 just as you got (but NOT "m/s" as you write!), though I think I did less work to get it!

Finally, your equation "D = 1/2 AT^2 2*2.53*t^2 = 1.26 secoinds" is mystifying! No, D (a distance) is not 1.26 seconds! I THINK what you did is solve the equation
(1/2) (2.53)T^2= 2 to get T= &radic:(2*2/2.53)= 1.26 seconds, but that is certainly not what you said!

Now that you know the time the box was sliding and its acceleration,
v= aT= (2.53 m/s2)(1.26 s)= 3.18 m/s.

You were right! (Which, considering how you wrote this out, is something of a miracle!)

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