Find initial velocity using energy

AI Thread Summary
A 1.4 kg box sliding 15m with a coefficient of kinetic friction of 0.3 experiences a friction force resulting in an energy loss of approximately 61.74J. The initial velocity of the box was calculated using energy principles, yielding a result of about 9.39 m/s, which aligns with other calculations. Discussions highlighted the importance of significant digits, noting that the coefficient of friction should only be reported with one significant digit, affecting the final answer. Participants emphasized the need for accuracy in calculations and the potential for losing points due to improper significant figures. The conversation reflects a thorough exploration of energy methods and friction in physics problems.
physics333
Messages
5
Reaction score
0
A 1.4 kg box slides 15m across the floor. The coefficient of kinetic friction between the box and the floor is 0.3. I found that friction force in slowing the box was -61.74J. Is this right?

What was the initial velocity of the box?
I worked this problem using Kf = Ki + Wnc and found that Vo =9.4m/s. Is this right? Also I would like to know how to work this problem another way with Energy. Any help with what formula to use or any help at all will be great.
 
Physics news on Phys.org
Would anyone else like to take a guess?
 
Well, this isn't an energy method but
μk=0.3 means that a=0.3g or about 2.94m/s2 so you could use that to get an answer. (9.39 m/s)
 
Originally posted by physics333
A 1.4 kg box slides 15m across the floor. The coefficient of kinetic friction between the box and the floor is 0.3. I found that friction force in slowing the box was -61.74J. Is this right?

What was the initial velocity of the box?

force of friction = (0.3)(1.4)(9.8)
force of friction = 4.116N

energy lost to friction = Fd
energy loss to friction = (4.116)(15)
energy loss to friction = 61.74J

E = (1/2)mv^2
v^2 = 2E/m
v = sqrt(2E/m)
v = sqrt(2(61.74)/(1.4))
v = 9.39m/s

all work shown and it agrees with the previous guy's answer
 
Originally posted by physics333
A 1.4 kg box slides 15m across the floor. The coefficient of kinetic friction between the box and the floor is 0.3. I found that friction force in slowing the box was -61.74J. Is this right?

You lose points for and incorrect number of significant digits.

Correct answer 62J
What was the initial velocity of the box?
I worked this problem using Kf = Ki + Wnc and found that Vo =9.4m/s. Is this right? Also I would like to know how to work this problem another way with Energy. Any help with what formula to use or any help at all will be great.

Likewise other responders there is at most 2 good digits in the given information, you cannot give more then that in the answer. A nit picky detail but one you need to be aware of. The common use of calcultors for the most trivial of computations makes this sort of error very common.
 


Originally posted by Integral
You lose points for and incorrect number of significant digits.

Correct answer 62J

If you're going to nitpick, do so correctly. The coefficient of kinetic friction has only one significant digit.
 
If you are an engineer, you either have the number of digits the question gave or use 3; whichever is greater. If the number starts with a 1, you use at least 4. Angles are to be expressed to at least 2 decimal places.
 
Last edited:
Obviously I am not an engineer.

You know I looked right past that single digit in μ
 


Originally posted by NateTG
If you're going to nitpick, do so correctly. The coefficient of kinetic friction has only one significant digit.
Ouch!

- Warren
 
Back
Top