A small block of mass m = 1.3 kg slides, without friction, along the loop-the-loop track shown. The block starts from the point P a distance h = 54.0 m above the bottom of the loop of radius R = 19.0 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?

1.

PE(B)=KE+PE(A)
mgh= KE+mg2R
KE= mgh- mg2R=
=mg(h-2R)

2.
mgh= mv²/2+mg2R
v²= 2g(h-2R)
a= v²/R= 2g(h-2R)/R
3.
mgh= mv²/2+mg2R
h= v²/2g +2R

Why did the block go to therapy? Because it had issues with its downward acceleration!

To find the kinetic energy of the mass at point A, we can use the conservation of mechanical energy. At point A, the block only has kinetic energy, so we can equate the initial potential energy (mgh) to the final kinetic energy (0.5mv^2).

Since the block starts from point P, which is a height of h = 54.0 m above the bottom of the loop, the initial potential energy is mgh = (1.3 kg)(9.8 m/s^2)(54.0 m).

To find the final velocity (v) at point A, we need to calculate the gravitational potential energy at that point. The potential energy at the bottom of the loop is 0, so the potential energy at point A is the initial potential energy minus the change in potential energy due to the height difference between P and A.

Using the formula for gravitational potential energy, ΔPE = mgh, the change in potential energy between P and A is mgh - mgh = 0. Therefore, the potential energy at point A is also mgh.

Setting the initial potential energy equal to the final kinetic energy, we have:

mgh = 0.5mv^2

Canceling out the mass:

gh = 0.5v^2

Solving for v:

v = √(2gh)

Plugging in the values:

v = √(2(9.8 m/s^2)(54.0 m))

v ≈ 30.97 m/s

Now, the kinetic energy at point A is:

KE = 0.5mv^2

KE = 0.5(1.3 kg)(30.97 m/s)^2

KE ≈ 603.72 J

So, the kinetic energy of the mass at point A is approximately 603.72 Joules!

As for the downward acceleration of the mass at point A, it can be calculated using centripetal acceleration. At the topmost point of the loop, the gravitational force provides the centripetal force, so we have:

mg = mv^2/R

Canceling out the mass:

g = v^2/R

Plugging in the values:

g = (30.97 m/s)^2 / 19.0 m

g ≈ 50.29 m/s^2

Therefore, the downward acceleration of the mass at point A is approximately 50.29 m/s^2!

To find the minimum height h for which the block will reach point A without leaving the track, we can use the concept of conservation of mechanical energy again. At the topmost point of the loop, the block should still have enough energy to reach point A. Therefore, we can equate the initial potential energy at point P to the final potential energy at point A:

mgh = 0.5mv^2

Canceling out the mass:

gh = 0.5v^2

Solving for h:

h = 0.5v^2/g

Plugging in the values:

h = 0.5(30.97 m/s)^2 / 9.8 m/s^2

h ≈ 50.17 m

Therefore, the minimum height h for the block to reach point A without leaving the track is approximately 50.17 meters. Make sure the block doesn't get a big head knowing this!

To find the kinetic energy of the block at point A on the loop, we can use the conservation of energy principle. At point P, the block has potential energy equal to mgh, where g is the acceleration due to gravity (approximately 9.8 m/s²). At point A, all of the potential energy is converted into kinetic energy.

So, the kinetic energy at point A is equal to the potential energy at point P:

KE_A = mgh

Substituting the given values, we have:

KE_A = (1.3 kg) * (9.8 m/s²) * (54.0 m)
= 663.36 Joules

Therefore, the kinetic energy of the mass at the point A on the loop is 663.36 Joules.

To find the downward acceleration of the mass at point A, we can use the centripetal acceleration formula:

a = v² / r

Here, v is the velocity at point A, and r is the radius of the loop.

At the highest point of the loop, the velocity will be zero as the block momentarily loses contact. Using circular motion principles, we know that at the top of the loop, the block experiences centripetal acceleration equal to g (the acceleration due to gravity). So, at point A, the centripetal acceleration is also g.

Therefore, the downward acceleration of the mass at the point A of the loop is approximately 9.8 m/s².

To find the minimum height h for the block to reach point A on the loop without leaving the track, we need to consider the minimum velocity required at point A. When the block is at the top of the loop, it must have enough velocity to overcome gravity and maintain contact with the track.

Using the principle of conservation of energy again, we can equate the potential energy at point P to the sum of kinetic and potential energy at point A:

mgh = (1/2)mv² + mgh_A

At the top of the loop, the block is momentarily at rest, so the kinetic energy is zero. Therefore:

mgh = mgh_A

Simplifying:

h = h_A

So, the minimum height required for the block to reach point A on the loop without leaving the track is 54.0 m.

To find the kinetic energy of the mass at point A on the loop, we need to determine its speed first. We can do this by using the principle of conservation of energy.

At point P, the block only has potential energy (mgh) because it is at a height h above the bottom of the loop and there is no friction. As the block moves down, it loses potential energy and gains kinetic energy. At point A, the block has lost all of its initial potential energy and gained an equal amount of kinetic energy.

To find the speed at point A, we can equate the potential energy at point P to the kinetic energy at point A.

mgh = (1/2)mv^2

Simplifying the expression, we get:

v^2 = 2gh

Taking the square root of both sides of the equation, we find:

v = sqrt(2gh)

To find the kinetic energy (KE) at point A, we can substitute the value of v in the expression:

KE = (1/2)mv^2 = (1/2)m(2gh) = mgh

Therefore, the kinetic energy of the mass at point A on the loop is mgh.

Now, let's calculate the downward acceleration of the mass at point A on the loop. We can use the centripetal acceleration equation:

a = v^2 / R

Since the block is at the loop's minimum point, we can substitute v = sqrt(2gh) and R = 19.0 m into the equation:

a = (2gh) / R

Plugging in the given values, we have:

a = (2 * 9.8 m/s^2 * h) / 19.0 m

Simplifying the expression, we get:

a = (19.6 * h) / 19.0

Now, let's find the minimum height h for which the block will reach point A on the loop without leaving the track. At point A, the centripetal acceleration needs to be greater than or equal to the acceleration due to gravity to keep the block on the loop, which means:

(19.6 * h) / 19.0 ≥ 9.8 m/s^2

Solving this inequality for h, we get:

h ≥ (9.8 m/s^2 * 19.0 m) / 19.6

Simplifying the expression, we find:

h ≥ 9.5 m

Therefore, the minimum height required for the block to reach point A on the loop without leaving the track is 9.5 meters.