What happens to kinetic energy of a snowball as it rolls across the lawn and gains mass
2 answers:
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Answer:
the equivalent mass :
the equation of the motion of the block of mass in terms of its displacement is =
Explanation:
Let use m₁ to represent the mass of the block and m₂ to represent the mass of the cylinder
The radius of the cylinder be = R
The distance between the center of the pulley to center of the block to be = x
Also, the angles of inclinations of the cylinder and the block with respect to the ground to be and
respectively.
The velocity of the block to be = v
The equivalent mass of the system =
In the terms of the equivalent mass, the kinetic energy of the system can be written as:
--------------- equation (1)
The angular velocity of the cylinder = : &
The inertia of the cylinder about its center to be = I
The angular velocity of the cylinder can be written as:
The kinetic energy of the system in terms of individual mass can be written as:
By replacing with
; we have:
------------------ equation (2)
Equating both equation (1) and (2); we have:
Therefore, the equivalent mass : which is read as;
The equivalent mass is equal to the mass of the block plus the mass of the cylinder plus the inertia by the square of the radius.
The expression for the force acting on equivalent mass due to the block is as follows:
Also; The expression for the force acting on equivalent mass due to the cylinder is as follows:
Equating the above both equations; we have the equation of motion of the equivalent system to be
which can be written as follows from the previous derivations
Finally; the equation of the motion of the block of mass in terms of its displacement is =
Answer:
y = 77.74 10⁻⁵ m
Explanation:
For this exercise we can use Newton's second law
F = m a
a = F / m
a = 4.9 10⁻¹⁶ / 9.1 10⁻³¹
a = 0.538 10¹⁵ m / s
This is the vertical acceleration of the electron.
Now let's use kinematics to find the time it takes to move the
x= 29 mm = 29 10⁻³ m
On the x axis
v = x / t
t = x / v
t = 29 10⁻³ / 1.7 10⁷
t = 17 10⁻¹⁰ s
Now we can look for vertical distance at this time.
y = t + ½ a t²
y = 0 + ½ 0.538 10¹⁵ (17 10⁻¹⁰)²
y = 77.74 10⁻⁵ m