there is no angle of inclination specified
kinetic energy becomes gravitational potential
1/2 m v^2 = m g h
Coasting along at 7.6m/s, a 60.0kg bicyclist on a 7.00kg bicycle encounters a small hill. If the she coasts uphill until coming to a stop, how high up the hill did she travel? (b) What height would the bike reach if she had gone to the store and had 5.50kg of groceries before coasting up the hill ?
I did Up the hill means force =2mg Sin Theta
And
Force fraction= forcegravity down the hill = mg*Sin Thera
Am I doing right ?
2 answers
a. KE = 0.5M*V^2 = 0.5*67*7.6^2 = 1935 J.
Work = Change in KE = 1935 J. (It changes to zero).
Work = F*d = Mg*d = 1935
67*9.8*d = 1935
d = 2.95 m up hill.
b. Same procedure as part a except M = 67 + 5.5 = 72.5 kg.
Work = Change in KE = 1935 J. (It changes to zero).
Work = F*d = Mg*d = 1935
67*9.8*d = 1935
d = 2.95 m up hill.
b. Same procedure as part a except M = 67 + 5.5 = 72.5 kg.
1 answer