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.
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.
4 answers
NOTE: The following calculations were not used: Wb, Fb, Fp, Fv.
y did u ruin my life
Hey Henry, just a friendly reminder: no one will understand your solutions if you don't explain your variables or your calculations