A 2.0 kg wood block is launched up awooden ramp that is inclined at a 35 degree

Question

A 2.0 kg wood block is launched up awooden ramp that is inclined at a 35 degree angle. Theblock’s initial speed is 10 m/s .

What vertical height does the block reach above its startingpoint?
Express your answer using two significant figures.

What speed does it have when it slides back down to its startingpoint?
Express your answer using two significant figures. A 2.0 kg wood block is launched up awooden ramp that is inclined at a 35 degree angle. Theblock’s initial speed is 10 m/s .

What vertical height does the block reach above its startingpoint?
Express your answer using two significant figures.

What speed does it have when it slides back down to its startingpoint?
Express your answer using two significant figures.

Explanation / Answer

For the vertical component: -- v(i)Sin -- To find the height it reaches: -- v(f)^2 = v(i)^2 + 2 ad - solve for d: d = [v(f)^2- v(i)^2] / 2a --> v(f) = 0 at maxheight d = (10Sin35)^2 / (2)(9.8) d = 1.7 m -- -- The speed it has when sliding back to the starting point isthe same as the initial speed in part a (10m/s), since in oneinstance kinetic energy is converted to potential and for thesecond part, potential energy is converted to kinetic energy. -- -- Hope this helps.

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