A 3kg mass has a speed of 5.0m/s at the bottom of a 37 degree incline. How far up the incline does the mass slide before it stops if the friction force opposing its motion is 20N?
If you draw a diagram it would be clearer, but I'll try to explain it
Initial velocity u=0, since it is dropped down an incline (I assume, usually is)
Final velocity v=5m/s (given)
We are given m=3kg. We can use this to deduce the weight of the mass. W=mg=3*10=30N
The weight is a downward force acting on the object that causes the mass to accelerate down the incline. However, the object is sliding down an incline of 37deg. That means that we have to split W into its two components, to find the one responsible for the object's acceleration down the incline.
W=W'sin37 where W' is the relevant component we need (draw it out to see)
W'=W/sin37 = 49.85kgms^-2
Since there is friction, we have to subtract it from W', take W'-20
Next we want to find the length of the incline the mass has travelled. We'll use one of the equations of kinematics.
S=(v^2-u^2)/2a is the most suitable.
S=5^2/[2(W'-20)]
Then you get your answer :)
*I haven't done this in awhile, please forgive any errors :)