Calculate the speed when the box reaches the bottom of the incline

AI Thread Summary
A 5kg box slides down a frictionless incline at a 15-degree angle over a length of 2.00 meters, starting from rest. The acceleration is calculated as 2.53 m/s² using the formula A = F/m, where the force is derived from the component of weight along the incline. The time taken to slide down is determined to be approximately 1.26 seconds. The final speed at the bottom of the incline is calculated to be 3.18 m/s. The calculations are confirmed to be correct, emphasizing the importance of proper unit notation and clarity in labeling axes.
cowgiljl
Messages
63
Reaction score
1
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?
 
Physics news on Phys.org
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.)
 
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!)
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top