A 0.21 -kg apple falls from a tree to the ground, 4.0 m below.Ignoring air resis
ID: 1742047 • Letter: A
Question
A 0.21 -kg apple falls from a tree to the ground, 4.0 m below.Ignoring air resistance, determine (a) the apple's kinetic energyand (b) its potential energy when the apples height is 1.0 m fromthe ground. Take ground level to be h=0 show step by step solution.. lifeavor pts. A 0.21 -kg apple falls from a tree to the ground, 4.0 m below.Ignoring air resistance, determine (a) the apple's kinetic energyand (b) its potential energy when the apples height is 1.0 m fromthe ground. Take ground level to be h=0 show step by step solution.. lifeavor pts.Explanation / Answer
A. The Kinetic Energy equation is KE=1/2*m*v2 where m isthe mass of the object, and v is the velocity of the object m=.21 kg Solve for v by using the equationv2=vo2+2a(x-xo) wherevo is the initial velocity, v is the final velocity, ais the acceleration due to gravity: 9.81 m/s2, and x isthe final position and xo is the initial position. Thereis no initial velocity (the apple falls from rest) so solving forthe final velocity: v2=2*9.81*(4-1)=58.86. Square rootboth sides v2=58.86 v=7.67 m/s. Wetake down to be the positive direction, so the even though it lookslike x final is 1m and x initial is 4m, we have to take thevelocity and position in the same direction (or we would have-58.86 which is undefined). Plug the velocity and mass into the KE equation1/2*.21*7.672= 6.17 J B. The gravitational potential energy equation is U(y)=m*g*h wherem is the mass, g is the acceleration due to gravity, and h is theheight above the earths surface. m=.21 kg, g=9.81m/s2, and h=1m U(y)= .21*9.81*1 =2.06 J Hope this is helps.
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