Where did this formula come from?

[XwakeriderX]

Homework Statement

In the attachment problem i was able to find the final velocity, from there someone helped me and got a formula that really confuses me. μ(m + M)g *x = 1/2(m + M) v^2

Homework Equations

All the work is write and the answer i just don't understand how they got
μ(m + M)g *x = 1/2(m + M) v^2

The Attempt at a Solution

The mass of the bulet m = 125g = 0.125kg

the mass of the block M = 5kg

the speed of the bullet vi = 300m/s

from law of conservation of momentum

mvi = (m + M)vf

then vf = mvi / (m + M)

= (0.125)(300)/(0.125 + 5)

= 7.32 m/s

Now from work energy theorem

W = ΔK

μ(m + M)g *x = 1/2(m + M) v^2

therefore the distance traveled


x = v^2 / μg = (7.32)^2 / (0.25)(9.8)(2) = 10.9 m

 

[PhanthomJay]

Homework Statement

In the attachment problem i was able to find the final velocity, from there someone helped me and got a formula that really confuses me. μ(m + M)g *x = 1/2(m + M) v^2

Homework Equations

All the work is write and the answer i just don't understand how they got
μ(m + M)g *x = 1/2(m + M) v^2

The equation comes from

Now from work energy theorem

W = ΔK

μ(m + M)g *x = 1/2(m + M) v^2

therefore the distance traveled


x = v^2 / μg = (7.32)^2 / (0.25)(9.8)(2) = 10.9 m

Are you familiar with the work energy theorem, which states that the total work done on an object is equal to its change in kinetic energy? If not, you can use Newton's 2nd law to find the acceleration, and then use the kinematic motion equations to find the distance traveled.

 

[XwakeriderX]

Hmm i found that W=FD and Fk=μkmg therefore W=μkmgD
Now i used W=(1/2)mv^2 then set them equal to each other then solved for D. It worked for me i hope I am not breaking any rules! I'll also try your way! Thanks!