Question

If a `1/2 kg` object moving at `7/4 m/s` slows to a halt after moving `3/8 m`, what is the coefficient of kinetic friction of the surface that the object was moving over?

Answer 1

The answer is `mu_k=1.25`

Explanation:

We solve in the horizontal direction `rarr^+`

`u=7/4ms^-2`

`v=0`

`s=3/8m`

`a=` acceleration

We use the equation

`v^2=u^2+2as`

`0=49/16+2*3/8*a`

`a=-49/16*4/3=-49/12ms^-2`

By Newton`s second Law

`F_r=ma=-49/4*1/2=-49/8N`

The reaction is `N=1/2g=1/2*9.8=4.9N`

The coefficient of kinetic friction is `mu_k=F_r/N=49/8*1/4.9=1.25`