Part 5 of 5 - Hitting the Ground Next, calculate the same values as the box slides along the top of the hill. These values are presented in the same order in each part, but you may need to find them in a different order. (a) What is the kinetic energy of the box at the starting point? K2=1531.64 J (b) What is the gravitational potential energy of the box at the starting point? U₂=0J (c) What is the total energy of the box at the starting point? E2 1531.64 J II. Velocity Now, use this to find the velocity of the box at the top of the hill. (a) What is the velocity of the box at the top of the hill? W 11.8 m/s 11.8 m/s (0.5 × ŷ + x +0.5 × 2) 2

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Sliding Up a Hill
When a problem contains only conservative forces, it is possible to model much of the dynamics using only energy conservation. In this application, you will return to the sliding box problem, but the ramp is replaced with a
strangely-shaped but frictionless slope. This makes this into a very difficult problem to solve kinematically, but straight-forward using energy conservation.
ū
h
You push a box of mass 22 kg with your car up to an icy hill slope of irregular shape to a height 5.2 m and width 4.5 m. The box has a speed 11.8 m/s when it starts up the hill, the same time that you brake. It then rises up to
the top (with no friction), which flattens off levelly for a distance 5 m. It ends at a sheer cliff, and then falls to the ground at the same level as where the box started (with no drag).
Use the ground as the zero for potential energies.
Part 5 of 5 - Hitting the Ground
Next, calculate the same values as the box slides along the top of the hill. These values are presented in the same order in each part, but you may need to find them in a different order. (a) What is the kinetic energy of the box
at the starting point?
K2=1531.64 J
(b) What is the gravitational potential energy of the box at the starting point?
U₂ =0J
(c) What is the total energy of the box at the starting point?
E2 = 1531.64 J
II. Velocity
Now, use this to find the velocity of the box at the top of the hill.
(a) What is the velocity of the box at the top of the hill?
-11.8 m/s
11.8 m/s
(0.5
128 Ý +0.5
× 2)
2
Transcribed Image Text:Sliding Up a Hill When a problem contains only conservative forces, it is possible to model much of the dynamics using only energy conservation. In this application, you will return to the sliding box problem, but the ramp is replaced with a strangely-shaped but frictionless slope. This makes this into a very difficult problem to solve kinematically, but straight-forward using energy conservation. ū h You push a box of mass 22 kg with your car up to an icy hill slope of irregular shape to a height 5.2 m and width 4.5 m. The box has a speed 11.8 m/s when it starts up the hill, the same time that you brake. It then rises up to the top (with no friction), which flattens off levelly for a distance 5 m. It ends at a sheer cliff, and then falls to the ground at the same level as where the box started (with no drag). Use the ground as the zero for potential energies. Part 5 of 5 - Hitting the Ground Next, calculate the same values as the box slides along the top of the hill. These values are presented in the same order in each part, but you may need to find them in a different order. (a) What is the kinetic energy of the box at the starting point? K2=1531.64 J (b) What is the gravitational potential energy of the box at the starting point? U₂ =0J (c) What is the total energy of the box at the starting point? E2 = 1531.64 J II. Velocity Now, use this to find the velocity of the box at the top of the hill. (a) What is the velocity of the box at the top of the hill? -11.8 m/s 11.8 m/s (0.5 128 Ý +0.5 × 2) 2
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