A small block of mass m=1.9kg slides, without friction, along the loop-the-loop track shown. The block starts from the point P a distance h=52.0m above the bottom of the loop of radius R=19.0m. What is the kinetic ennergy of the mass at the point A on the loop? What is the downward accelerations of the mass at point A? What is the minimum height h for which the block will reach point A on the loop without leaving the track?
To find the kinetic energy of the mass at point A on the loop, we need to apply the principle of conservation of energy.
First, let's analyze the situation at point P. The block has gravitational potential energy (PE) due to its height above the bottom of the loop.
PE at P = m * g * h
where m is the mass (1.9kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height (52.0m).
Next, let's consider the situation at point A on the loop. At this point, all the potential energy has been converted into kinetic energy (KE) since there is no friction.
KE at A = PE at P
Using the equation for potential energy at P, we can determine the kinetic energy at A:
KE at A = m * g * h
Now, let's calculate the value:
KE at A = 1.9kg * 9.8 m/s^2 * 52.0m = 964.8 Joules
So, the kinetic energy of the mass at point A is 964.8 Joules.
To find the downward acceleration of the mass at point A, we can use the centripetal acceleration equation:
a = v^2 / r
where a is the acceleration, v is the velocity, and r is the radius of the loop.
At point A, the velocity is tangent to the circle, so we can use the equation for the velocity of an object at the bottom of a loop-the-loop:
v = √(g * r)
Substituting the values:
v = √(9.8 m/s^2 * 19.0m) = 9.8 m/s
Now, we can calculate the downward acceleration:
a = (9.8 m/s)^2 / 19.0m = 5.047 m/s^2
So, the downward acceleration of the mass at point A is 5.047 m/s^2.
Finally, let's determine the minimum height (h) for the block to reach point A on the loop without leaving the track.
To reach point A on the loop, the minimum height (h_min) needs to be such that the block maintains sufficient speed (velocity) throughout the loop.
Using the equation for velocity at the bottom of the loop mentioned earlier, we have:
v = √(g * r)
Rearranging the equation to solve for g:
g = v^2 / r
At the top of the loop, the velocity (v) is zero, so we can calculate the minimum value for g:
g_min = (0 m/s)^2 / 19.0m = 0 m/s^2
Now, we can calculate the minimum height (h_min) using the potential energy equation:
PE at A = m * g_min * h_min
Since g_min is zero, PE at A will also be zero.
Therefore, the minimum height (h_min) for the block to reach point A on the loop without leaving the track is zero.