Pre-Lab Q1: Describe the kinetic and potential energy at position 0, the instant

Question

Pre-Lab Q1: Describe the kinetic and potential energy at position 0, the instant when the ball is first released from rest:

a) Kinetic energy at position 0: i. maximum ii. increasing iii. decreasing iv. zero

b) Potential energy at position 0: i. maximum ii. increasing iii. decreasing iv. zero

Pre-Lab Q2: Describe the kinetic and potential energy at position 1, when the height is half of the original and the ball is on its way toward the floor:

a) Kinetic energy at position 0: i. maximum ii. increasing iii. decreasing iv. zero

b) Potential energy at position 0: i. maximum ii. increasing iii. decreasing iv. zero

Pre-Lab Q3: Describe the kinetic and potential energy at position 3, the instant JUST BEFORE the ball touches the ground (the position or height is approximately zero):

a) Kinetic energy at position 3: i. maximum ii. increasing iii. decreasing iv. zero

b) Potential energy at position 3: i. maximum ii. increasing iii. decreasing iv. zero

Pre-Lab Q4: Describe the kinetic and potential energy at position 4, the instant JUST AFTER the ball leaves the ground (the position or height is approximately zero):

a) Kinetic energy at position 4: i. maximum ii. increasing iii. decreasing iv. zero

b) Potential energy at position 4: i. maximum ii. increasing iii. decreasing iv. zero

Pre-Lab Q5: Since there is minimal air resistance acting on the ball (little energy out), how are the changes in gravitational potential energy related to the change in kinetic energy?

Explanation / Answer

Kinetic energy is given by:

KE = 0.5*m*V^2

Potential energy is given by:

PE = m*g*h

Q1:

When Ball is released from rest

Vi = 0, So KE = Zero

h = Hmax, So PE = m*g*Hmax = Mximum

Q2

When height is half the original

V = Vi + a*t = 0 + g*t

So with time V is increasing, which means KE is increasing

PE = m*g*h, Since h is decreasing, So PE is decreasing

Q3

Just before ball touches the ground

V = Vmax, So KE = maximum

h = 0, So PE = Zero

Q4

V = Vmax + a*t

a = -g, So

V = Vmax - g*t, which means Velocity is decreasing, So KE is decreasing

PE = mgh

Since h is increasing, So PE is Also increasing

Q5.

Since there is minimal air resistance, So Energy is conserved, in other words

Change in PE = Chnage in KE

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