A block of mass 0.78kg starts from rest at point A and slides down a frictionless hill of height h. At the bottom of the hill it slides across a horizontal piece of track where the coefficient of kinetic friction is 0.29. This section (from points B to C) is 4.08m in length. The block then enters a frictionless loop of radius r= 2.87m. Point D is the highest point in the loop. The loop has a total height of 2r. Note that the drawing below is not to scale.

a)What is the minimum speed of the block at point D that still allows the block to complete the loop without leaving the track?
b)What is the minimum kinetic energy for the block at point C in order to have enough speed at point D that the block will not leave the track?
c)What is the minimum kinetic energy for the block at point B in order to have enough speed at point D that the block will not leave the track?
d)What is the minimum height from which the block should start in order to have enough speed at point D that the block will not leave the track?

(a) At the point D

ma=mg
mv²/r=mg
v=sqrt(g•r)=sqrt(9.8•2.87)=5.3 m/s
(b)
KE(C)=PE+KE=
=mg•2r+mv²/2=
=m(2gr+5.3²/2)=54.8 J.

(c)
KE(B)=KE(C) + W(fr)=
= KE(C) + μ•mg•s=
=54.8+0.29•0.78•9.8•4.08=
=63.8 J

(d)
PE1=KE(B)
mgh= KE(B)
h=KE(B)/mg=63.8/0.78•9.8 =8.35 m

a) Why did the block become so competitive all of a sudden? It's trying to complete the loop without leaving the track! Well, to achieve this, we need to find the minimum speed at point D. To solve this, we can use the conservation of mechanical energy. The potential energy at point A is equal to the sum of the kinetic energy at point D and the potential energy at point D. Let's set up the equation and solve for the minimum speed:

mgh = 1/2 mv^2 + mg(2r)

Where m is the mass of the block, g is the acceleration due to gravity, h is the height of the hill, r is the radius of the loop, and v is the speed at point D.

b) Now, let's talk about the minimum kinetic energy for the block at point C. We want the block to have enough speed at point D so that it doesn't leave the track. The kinetic energy at point C should be equal to the potential energy at point D. Let's set up the equation and solve for the minimum kinetic energy:

1/2 mv_c^2 = mg(2r)

Where m is the mass of the block, g is the acceleration due to gravity, v_c is the speed at point C, and r is the radius of the loop.

c) Moving on to the minimum kinetic energy at point B. Again, we want the block to have enough speed at point D to stay on the track. The kinetic energy at point B should be equal to the potential energy at point D. Let's set up the equation and solve for the minimum kinetic energy:

1/2 mv_b^2 = mg(2r)

Where m is the mass of the block, g is the acceleration due to gravity, v_b is the speed at point B, and r is the radius of the loop.

d) Lastly, let's calculate the minimum height from which the block should start to achieve enough speed at point D. We can use the same conservation of mechanical energy equation as in part a, but this time we solve for height:

h = 1/2 v_d^2/(2g)

Where h is the height of the hill, v_d is the speed at point D, and g is the acceleration due to gravity.

I hope these calculations help the block fulfill its dream of completing the loop without leaving the track!

To solve this problem, we can use conservation of energy.

a) The minimum speed of the block at point D can be found by equating the potential energy at point D with the sum of the kinetic energy and potential energy at point C.

At point D:
Potential energy = mgh = m(2r)g

At point C:
Potential energy = mgh + 0.5mv^2 (kinetic energy)

Since the block is at its minimum speed, it will have just enough speed to reach the top of the loop without leaving the track. At the top of the loop (point D), the block is momentarily at rest.

Therefore, potential energy at point C = potential energy at point D

mgh + 0.5mv^2 = m(2r)g

Simplifying and solving for v, we have:
v^2 = 2gh(2r - h)
v = sqrt(2gh(2r - h))

b) The minimum kinetic energy for the block at point C can be found by setting the sum of the kinetic energy and potential energy at point C equal to the sum of the kinetic energy and potential energy at point D, since the block is at its minimum speed.

Potential energy at point C = mgh + 0.5mv^2
Potential energy at point D = m(2r)g

Since the block is at its minimum speed, the kinetic energy at point C is zero.

mgh + 0.5mv^2 = m(2r)g

Simplifying and solving for KEc (kinetic energy at point C):
KEc = m(2r)g - mgh
KEc = m(2r - h)g

c) The minimum kinetic energy for the block at point B can be found by considering the friction between the block and the horizontal track.

The work done by friction is equal to the change in kinetic energy. The work done by friction is given by the equation:

Work = force of friction * distance = μ(mg) * distance

where μ is the coefficient of kinetic friction.

Since the block is at its minimum speed, the kinetic energy at point B is zero.

Therefore, μ(mg) * distance = 0
μ(mg) * distance = KEb

d) The minimum height from which the block should start can be found by equating the potential energy at point A with the sum of the kinetic energy and potential energy at point D.

Potential energy at point A = mgh
Potential energy at point D = m(2r)g

Since the block is at its minimum speed, the kinetic energy at point A is zero.

mgh = m(2r)g

Simplifying and solving for h:
h = 2r

Therefore, the minimum height should be equal to 2 times the radius of the loop.

To solve this problem, we need to consider the conservation of energy and the forces acting on the block at different points.

a) The minimum speed at point D can be determined by equating the centripetal force required for circular motion to the gravitational force acting on the block:

Centripetal force = Gravitational force

(mass x velocity^2) / radius = mass x acceleration due to gravity

Using this equation, we can solve for velocity:

velocity = sqrt(radius x acceleration due to gravity)

In this case, the radius is given as 2.87m and the acceleration due to gravity is approximately 9.8 m/s^2.

b) At point C, the block will experience both the gravitational force and the force of kinetic friction. We can find the minimum kinetic energy required by equating the gravitational force to the force of kinetic friction:

Gravitational force = Force of kinetic friction

mass x acceleration due to gravity = coefficient of kinetic friction x Normal force

Since the normal force is equal to the weight of the block (mass x acceleration due to gravity), we can rewrite the equation as:

mass x acceleration due to gravity = coefficient of kinetic friction x mass x acceleration due to gravity

This simplifies to:

acceleration due to gravity = coefficient of kinetic friction x acceleration due to gravity

Rearranging the equation, we find:

acceleration due to gravity = coefficient of kinetic friction x g

The minimum kinetic energy can be calculated using the formula:

Kinetic energy = (1/2) x mass x velocity^2

Given the coefficient of kinetic friction and the acceleration due to gravity, we can substitute these values into the equation and solve for the minimum kinetic energy.

c) At point B, there is no vertical force acting on the block, so the only force to consider is the force of kinetic friction. We can calculate the minimum kinetic energy at point B using the same formula as above, but with the distance traveled instead of the radius for the centripetal force.

d) To calculate the minimum height from which the block should start, we need to consider the conservation of mechanical energy. At point D, the total mechanical energy is equal to the sum of the potential energy at the starting point (point A) and the kinetic energy at point D.

Total Mechanical Energy = Potential Energy at A + Kinetic Energy at D

Potential Energy at A = mass x acceleration due to gravity x height

Substituting the expressions for the potential energy and kinetic energy, we can solve for the minimum height.