3.92 kg block located on a horizontal floor is pulled by a cord that exerts a force F = 11.5 N at an angle theta = 15.5o above the horizontal, as shown in the Figure. The coefficient of kinetic friction between the block and the floor is 0.09. What is the speed of the block 5.1s after it starts moving?

1 answer

Well, I will sketch out this one but then you try the next one.
force down on floor = 3.92 g - 11.5 sin 15.5
so
friction force = 0.09(3.92 g - 11.5 sin 15.5)
call it ff
then
11.5 cos 15.5 - ff = 3.92 a
calculate a
initial speed = 0
so
v = a t = a * 5.1 meters/second