A 3.5-kg block slides down a ramp with friction, as shown in the figure. The top of the ramp is 1.5 m above the ground; the bottom of the ramp is h = 0.17 m above the ground. The block leaves the ramp moving horizontally, and lands a horizontal distance d away. Find the distance d for the case in which friction on the ramp does -9.9 J of work on the block before it becomes airborne.

Question

_______m

Answer 1

How far down did it slide?
1.5 - .17 = 1.33 meters
change in potential energy = m g h
= 3.5 * 9.81 * 1.33 = 45.7 Joules
so ke at launch = 45.7-9.9 = 35.8 Joules
so
(1/2)(35) u^2 = 35.8
u = 1.43 m/s horizontal launch speed

Now it falls .17 meter
.17 = 4.9 t^2
t = .186 seconds
horizontal d = 1.43 * .186 = .267 meters or about 27 cm

Answer 2

Kinetic energy at bottom: initial energy+gpe added - friction work.

KE at bottom= 3.5*9.8*(1.5-.17)-9.9 Joules

velocity at launch:
1/2 m v^2= KE at bottom
solve for v, that is the horizontal velocity.

time in air: hf=hi-1/2 9.8 t^2
t= sqrt(2*.17/9.8) seconds
horizontal distance:
v*t= you do that calculation