Question

If a `2 kg` object moving at `5 m/s` slows down to a halt after moving `3 m`, what is the friction coefficient of the surface that the object was moving over?

Answer 1

`u_k~=0,425`

Explanation:

`E_k=1/2*m*v^2 " The kinetic energy of object"`
`W=F_f* Delta x " Work doing by friction force"`
`F_f=u_k*N " N: normal force to contacting surfaces"`
`"The kinetic energy turns work"`
`1/2*cancel(m)*v^2=u_k.cancel(m)*g*Delta x`
`1/2*5^2=u_k*9,81*3`
`25=u_k*2*3*9,81`
`u_k=25/(2.3.9,81)`
`u_k~=0,425`