Asked by senko

A 2.0kg wood block is launched up a wooden ramp that is inclined at a 26 angle. The block's initial speed is 8.0 m/s. The coefficient of kinetic friction of wood on wood is 0.200.

a) What vertical height does the block reach above its starting point?
Express your answer to two significant figures and include the appropriate units.

b) What speed does it have when it slides back down to its starting point?
Express your answer to two significant figures and include the appropriate units.

Please show steps! Thank you.
Answered by Henry
Wb = m*g = 2kg * 9.8N/kg = 19.6 N. = Wt.
of block.

Fb = 19.6N @ 26o. = Force of block.
Fp = 19.6*sin26 = 8.59 N. = Force parallel to ramp.
Fv = 19.6*cos26 = 17.62 N. = Force perpendicular to ramp.

a. Vo = 8 m/s.
Yo = 8*sin26 = 3.51 m/s.

hmax = (Y^2-Yo^2)/2g.
hmax = (0-12.3)/-19.6 = 0.627 m.

b. V^2 = Vo^2 + 2g*d.
V^2 = 0 + 19.6*0.627 = 12.29
V = 3.51 m/s.
Answered by Henry
NOTE: The following calculations were not used: Wb, Fb, Fp, Fv.
Answered by Anonymous
y did u ruin my life
Answered by Anon
Hey Henry, just a friendly reminder: no one will understand your solutions if you don't explain your variables or your calculations
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