Suppose you have a 113-kg wooden crate resting on a wood floor. (μk = 0.3 and μs = 0.5) (a) What maximum force (in N) can you ex
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This statement is false. Increasing the two objects' mass (I'm guessing) will actually increase their gravitational force. This is because of the equation:
If the distance was increased, then the statement would be true, but since you are increasing mass, which is proportional to the Force of Gravity, you are in fact, increasing the gravitational force between the two objects.
If the distance was increased, then the statement would be true, but since you are increasing mass, which is proportional to the Force of Gravity, you are in fact, increasing the gravitational force between the two objects.
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
Answer:
The components of the moving frame is (8.07c, -2, 3, 9.493)
Solution:
As per the question:
Velocity of moving frame w.r.t original frame 0.85c
Point 'a' of an event in one reference frame corresponds to the (x, y, z, t) coordinates of the plane
a = (0, - 2, 3, 5)
Now, according the the question, the coordinates of moving frame, say (X, Y, Z, t'):
New coordinates are given by:
X =
X =
X =
Now,
Y = y = - 2
Z = z = 3
Now,