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

In summary, a 5kg box slides down a frictionless plane that makes an angle of 15 degrees with the horizontal and has a length of 2.00 meters. If the box starts from rest, the acceleration can be calculated using the formula A=F/m, which results in 2.53m/s^2. The speed of the box when it reaches the bottom of the incline can be determined using the equation Vf^2 = Vi^2+2AD, which gives a final speed of 3.18m/s. The weight of the box can also be calculated using the formula w = mg, which is 49N. However, in this problem, the mass is irrelevant and the acceleration along the
  • #1
cowgiljl
63
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?
 
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  • #2
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.)
 
  • #3
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!)
 

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

What is the formula for calculating the speed at the bottom of an incline?

The formula for calculating the speed at the bottom of an incline is v = √(2gh), where v is the speed, g is the acceleration due to gravity, and h is the height of the incline.

How is the acceleration due to gravity related to the speed at the bottom of an incline?

The acceleration due to gravity affects the speed at the bottom of an incline because it determines how quickly an object will accelerate down the incline. The higher the acceleration due to gravity, the faster the object will accelerate and the higher the speed will be at the bottom of the incline.

Is the speed at the bottom of an incline affected by the mass of the object?

Yes, the speed at the bottom of an incline is affected by the mass of the object. The heavier the object, the more force is needed to accelerate it down the incline, resulting in a higher speed at the bottom.

What units should be used when calculating the speed at the bottom of an incline?

The units used for calculating the speed at the bottom of an incline are meters per second (m/s). These units are derived from the formula v = √(2gh), where g is measured in meters per second squared (m/s²) and h is measured in meters (m).

Can the speed at the bottom of an incline be greater than the speed at the top?

Yes, the speed at the bottom of an incline can be greater than the speed at the top. This is because the object gains kinetic energy as it accelerates down the incline, resulting in a higher speed at the bottom compared to the top.

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