A small block of mass m = 1.1 kg slides, without friction, along the loop-the-loop track shown. The block starts from the point P a distance h = 56.0 m above the bottom of the loop of radius R = 17.0 m. What is the kinetic energy of the mass at the point A on the loop?
What is the downward acceleration of the mass at the point A of the loop?
What is the minimum height h for which the block will reach point A on the loop without leaving the track?
Why did the block bring a parachute to the loop-the-loop track? Because it wanted to make sure it had a safe "landing" at point A! 😄
To find the kinetic energy at point A, we can use the conservation of energy principle. At point P, the block only has potential energy, which is converted into kinetic energy at point A.
The potential energy at point P is given by: P.E. = mgh
Substituting the given values, we have: P.E. = (1.1 kg)(9.8 m/s^2)(56.0 m)
The kinetic energy at point A is equal to the potential energy at point P, since there is no loss due to friction. So the kinetic energy at point A is also: K.E. = (1.1 kg)(9.8 m/s^2)(56.0 m)
Now let's move on to the downward acceleration at point A. Unlike my clown car, the block doesn't have an accelerator pedal. But it does experience a net force directed towards the center of the loop.
We can use the equation: F_net = m * a
The net force is provided by the gravitational force (mg) and the centripetal force (mv^2 / R), which are both directed downward at point A.
So we have: mg + mv^2 / R = m * a
Now, let's solve for the acceleration (a). Plug in the given values for mass (m) and radius (R): 1.1 kg and 17.0 m, respectively.
Finally, we come to the question of the minimum height (h) for the block to reach point A on the loop without leaving the track. It needs just the right amount of "fuel" (potential energy) to complete the loop without falling off.
To find the minimum height, we can equate the potential energy at the highest point (h = 0) to the kinetic energy at point A.
So we have: P.E. (highest point) = K.E. (point A)
Solving for h, we can substitute the given values for mass (m), radius (R), and the gravitational acceleration (g). This will give us the minimum height needed for the block to reach point A without leaving the track.
To find the kinetic energy of the mass at point A on the loop, we need to consider the conservation of mechanical energy. The mechanical energy of the system (consisting of the mass and the Earth) is conserved as there is no friction.
The mechanical energy of the system can be expressed as the sum of potential energy and kinetic energy:
E = PE + KE
At point A on the loop, the potential energy is zero since it is at the same height as the reference point. Therefore, the mechanical energy at point A is equal to the kinetic energy:
E_A = KE_A
Now, to find the kinetic energy at point A, we can use the conservation of mechanical energy. The mechanical energy at point P (starting point) is equal to the mechanical energy at point A:
E_P = E_A
The mechanical energy at point P is the sum of potential energy and kinetic energy at point P:
E_P = PE_P + KE_P
Since there is no friction, the mechanical energy is conserved, so:
PE_P + KE_P = KE_A
Now, let's solve for the kinetic energy at point A:
KE_A = PE_P + KE_P
First, we need to find the potential energy at point P. The potential energy at point P is given by:
PE_P = m * g * h
where m is the mass, g is the acceleration due to gravity, and h is the height above the bottom of the loop.
Given data:
m = 1.1 kg
h = 56.0 m
Substituting these values in, we get:
PE_P = 1.1 kg * 9.8 m/s^2 * 56.0 m
Next, we need to find the kinetic energy at point P. The kinetic energy at point P can be found using the equation:
KE_P = 0.5 * m * v_P^2
where v_P is the velocity at point P. Since we know that there is no friction, the total mechanical energy at point P is equal to the total mechanical energy at point A. This means that the velocity at point P is equal to the velocity at point A.
Substituting these values in, we get:
KE_P = 0.5 * 1.1 kg * v_A^2
Since KE_A = KE_P, we can equate the two expressions for kinetic energy:
PE_P + KE_P = KE_A
1.1 kg * 9.8 m/s^2 * 56.0 m + 0.5 * 1.1 kg * v_A^2 = KE_A
Now, let's move on to finding the downward acceleration of the mass at point A on the loop. We can use the circular motion equations to find the acceleration at this point.
The centripetal acceleration of an object moving in a circle of radius R at a speed v is given by:
a_c = v^2 / R
In this case, the centripetal acceleration at point A is also equal to the gravitational acceleration, as it provides the necessary centripetal force to keep the mass moving in a circle. Therefore, we have:
a_c = g
Now, let's find the minimum height h for which the block will reach point A on the loop without leaving the track. To determine this height, we need to consider the condition for the block to stay in contact with the loop at point A, which is that the normal force must be greater than or equal to zero.
At the highest point of the loop (point A), the net force is equal to the gravitational force acting downward:
F_net = m * g
The net force is the difference between the gravitational force and the normal force:
F_net = m * g - N
Since the normal force at point A is equal to zero (minimum height), we can set up the equation:
m * g - N = 0
Solving for the minimum height h, we find:
m * g - N = 0
m * g = N
N = m * g
Substituting this back into the equation, we get:
f_net = m * g - N
m * g = m * g - N
m * g = m * g - m * g
m * g = 0
Therefore, the minimum height h for which the block will reach point A on the loop without leaving the track is 0.