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Answer:
The gravitational potential energy of the nickel at the top of the monument is 8.29 J.
Explanation:
We can find the gravitational potential energy using the following formula.
Identifying given information.
The nickel has a mass , and it is a the top of Washington Monument.
The Washington Monument has a height of , thus we need to find the equivalence in meters using unit conversion in order to find the gravitational potential energy.
Converting from feet to meters.
Using the conversion factor 1 m = 3.28 ft, we have
That give u s
Finding Gravitational Potential Energy.
We can replace the height and mass on the formula
And we get
The gravitational potential energy of the nickel at the top of the monument is 8.29 J.
Answer:
the block reaches higher than the sphere
\frac{y_{sphere}} {y_block} = 5/7
Explanation:
We are going to solve this interesting problem
A) in this case a sphere rolls on the ramp, let's find the speed of the center of mass at the exit of the ramp
Let's use the concept of conservation of energy
starting point. At the top of the ramp
Em₀ = U = m g y₁
final point. At the exit of the ramp
Em_f = K + U = ½ m v² + ½ I w² + m g y₂
notice that we include the translational and rotational energy, we assume that the height of the exit ramp is y₂
energy is conserved
Em₀ = Em_f
m g y₁ = ½ m v² + ½ I w² + m g y₂
angular and linear velocity are related
v = w r
the moment of inertia of a sphere is
I = m r²
we substitute
m g (y₁ - y₂) = ½ m v² + ½ ( m r²) (
)²
m g h = ½ m v² (1 + )
where h is the difference in height between the two sides of the ramp
h = y₂ -y₁
mg h = (
m v²)
v = √5/7 √2gh
This is the exit velocity of the vertical movement of the sphere
v_sphere = 0.845 √2gh
B) is the same case, but for a box without friction
starting point
Em₀ = U = mg y₁
final point
Em_f = K + U = ½ m v² + m g y₂
Em₀ = Em_f
mg y₁ = ½ m v² + m g y₂
m g (y₁ -y₂) = ½ m v²
v = √2gh
this is the speed of the box
v_box = √2gh
to know which body reaches higher in the air we can use the kinematic relations
v² = v₀² - 2 g y
at the highest point v = 0
y = vo₀²/ 2g
for the sphere
y_sphere = 5/7 2gh / 2g
y_esfera = 5/7 h
for the block
y_block = 2gh / 2g
y_block = h
therefore the block reaches higher than the sphere