How far will a box slide with given initial velocity, friction, and mass?

[subopolois]

Homework Statement

How far will a box slide if its initial velocity is 1.5 m/s, kinetic friction is 0.15 and the object is 30 kg

Homework Equations

a= (kinetic friction)(acceleration due to gravity) and
V(final)^2 - V(initial)^2 = 2(a)(D)

The Attempt at a Solution

a= 0.15 x 9.81 m/s^2
= -1.47 m/s^2 (negative since it is slowing down)

D= V(final)^2 - V(initial)^2 / 2a
= 0 - (1.5 m/s)^2 / 2(-1.47 m/s^2)
= -2.25 m/s^2 / -2.94 m/s^2
= 0.76 m

did i do everything right? seems to me like it should go a little further given the value of friction...

 

[Doc Al]

Looks good to me!

 

[Doc Al]

You solved it using kinematics, which is perfectly fine. What other approach might you have used?

 

[subopolois]

Id say id be able to use kinetic energy to solve it, yay or nay?

 

[Doc Al]

Id say id be able to use kinetic energy to solve it, yay or nay?

Yes, you could use energy methods to solve this. Good!

(Try setting it up enough to convince yourself that you get the same equation in the end.)

 

[subopolois]

ok so let's say i use energy. I have this equation
v(initial)^2 = 2(mu)(g)(delta r)

and i rearrange to find r
r= v(initial)^2 / 2(mu)(g)

I put all my numbers in and i get a different answer, 0.54 meters. what did i do wrong?

 

[Doc Al]

ok so let's say i use energy. I have this equation
v(initial)^2 = 2(mu)(g)(delta r)

and i rearrange to find r
r= v(initial)^2 / 2(mu)(g)

Looks good. Note that this is exactly the same equation you ended up with when you solved it using kinematics.

I put all my numbers in and i get a different answer, 0.54 meters. what did i do wrong?

Check your arithmetic.

 

[subopolois]

ahhh. forgot to square the initial velocity. Thanks for all your help!