Question

A 10.0 kg block is released from rest down a 30° ramp.

It slides 6.0m along the ramp against a frictional force
of 16N. What is its speed after sliding 6.0m along the ramp?
*SOLVE USING ENERGY

Answer 1

6.298 m/s

Step-by-step explanation:

`F_(net) =ma`

`F_F=u_k*F_N` friction force equation

`ma=mgsin(x)-umgcos(x)`

We are given

`m=10\\g=9.81\\F_F=16\\V_i=0`

`umgcos(x)=F_F`

`u*10*9.81*cos(x)=16`

`u=0.18833036`

`ma=mgsin(x)-umgcos(x)`

`10*a=10*9.81*sin(30)-0.18833036*10*9.81*cos(30)`

`a=3.305`

`V_(f) ^(2) =V_(i) ^(2)+2ax`

`V_(f) ^(2) =0 ^(2)+2*3.305*6`

`V_f=6.298`