Question

Question #6e844

Answer 1

Given

`m-> "mass of the block"=9.25kg`

`F-> "horizontal force applied on the block"=55N`

`mu_k-> "coefficient of kinetic friction"=0.175`

`F_"fric"-> "frictional force resisting the block"`
`" "" "=mu_kmg=0.175xx9.25xx9.8N=15.86375N`

So net force acting on the block

`F_n=F-F_"fric"=(55-15.86375)=39.13625N`

So acceleration produced

`a=F_n/m=(39.13625N)/(9.25kg)~~4.23ms^-2`

If time taken to cover the distance `s=3m` be t sec then by using equation of kinematics we can write

`S=uxxt+1/2at^2`

`=>3=0xxt+1/2xx4.23xxt^2`

`t=sqrt(6/4.23)s=~~1.19s`