A 1.9 kg block slides down a ramp, as shown in the figure. (Figure 1) The ramp i

Question

A 1.9 kg block slides down a ramp, as shown in the figure. (Figure 1) The ramp in the figure is not frictionless. The top of the ramp is h1 = 1.5 m above the ground; the bottom of the ramp is h2 = 0.25 m above the ground. The block leaves the ramp moving horizontally, and lands a horizontal distance d away. Friction on the ramp does -9.7 J of work on the block before it becomes airborne. I've tried everything including looking at the other similar problems. I have one attempt left and just need an answer at this point. Proper would still be awesome!

Explanation / Answer

E = mgH

E =1.9*9.8*1.5

E =27.93 J

E1 = E - 9.7

E1 = 18.23 J is kinetic energy at h=0.25m

Beside’s we have E1=0.5*m*v2

Hence v= (2E1/m) is speed and this speed is horizontal

v = 19.1894737 m/s

time t=d/v to fall from h = 0.25

so

h = ½ g*t2

h = ½ *g*(d/19.1894737)2

0.25 = ½ *9.8*(d/19.1894737)2

(0.05102041)0.5 = d/19.1894737

d = 4.33 m

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