Question

If an object with a mass of `10 kg` is moving on a surface at `45 m/s` and slows to a halt after `6 s`, what is the friction coefficient of the surface?

Answer 1

`u_k=0,0051`

Explanation:

`v_l=v_i-a*t`
`v_l:"last velocity"`
`v_i:"initial velocity"`
`a:"acceleration"`
`t:"time"`

`v=v/2-a*6`
`6*a=v-v/2`
`6*a=-v/2`
`a=-v/12" "a=-3,75" " m/s^2" 'acceleration of object'"`

`v_l^2=v_i^2-2*a* Delta x`
`v^2=(v/2)^2-2.3,75*Delta x`

`v^2-v^2/4=7,50*Delta x`

`(3*v^2)/4=7,50*Delta x`
`3*v^2=30*Delta x`
`v^2=10*Delta x" ; "45^2=10*Delta x" ; "2025=10*Delta x`
`Delta x=202,5 " m"`
`E_k=1/2*m*v^2 "the kinetic energy of object"`
`W_f=F_f*Delta x" work doing by friction force"`
`F_f=u_k*m*g" "W_f=u_k*m*g*Delta x`
`E_k=W_f`
`1/2*cancel(m)*v^2=u_k*cancel(m)*g*Delta x`

`u_k=v^2/(2*g*Delta x)`

`u_k=45^2/(2*9,81*202,5)=2025/397305`

`u_k=0,0051`