A 0.2 kg object is released from rest at point A on a track that is one-quarter of a circle with radius 1.6 m, as shown in the figure. It slides down and reaches point B with a speed of 4.8 m/s. Then it slides on a level surface a distance of 3 m until it comes to rest at point C.

a) How much work is done by friction on the package at it slides down the circular track from A to B?
b) What is the coefficient of kinetic friction on the horizontal surface?
c) What average power is produced by friction as the object moved from B to C?
d) What instantaneous power is produced by friction at point B?

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W=mgh-1/2 mv^2=0.2×9.8×1.6-1/2×0.2×〖4.8〗^2=0.832 J

2.b
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F_Fric=μ F_n but F_n=mg
F_Fric=μ mg
And
F=ma
a=F/m=(μ mg)/m=μ g

μ=a/g……………..1
So
a=v^2/d=〖4.8〗^2/3=7.68 m/s^2
From ..1
μ=a/g=7.68/9.8=0.78

c.
P ̅=W/t=(F_Fric×d)/t=F_Fric×v ̅

but F_Fric=μ mg=0.78×0.2×9.8=1.53N and v ̅=4.8/2=2.4 m/s

∴ P ̅=F_Fric×v ̅=1.53×2.4=3.67 Watt

d.
P=F_Fric×v=1.53×4.8=7.34 Watt

a) To determine the work done by friction on the package as it slides down the circular track from point A to point B, we need to first calculate the gravitational potential energy difference between the two points.

The gravitational potential energy can be calculated using the formula:
PE = m * g * h

Where:
m = mass of the object = 0.2 kg
g = acceleration due to gravity = 9.8 m/s^2
h = height difference between the two points = radius of the track (1.6 m)

PE_A = 0.2 kg * 9.8 m/s^2 * 1.6 m = 3.136 J

The object reaches point B with a speed of 4.8 m/s. Since it comes to rest at point C, we can assume all of its kinetic energy at point B is converted into work done by friction.

Kinetic energy at point B can be calculated using the formula:
KE = 0.5 * m * v^2

Where:
m = mass of the object = 0.2 kg
v = velocity at point B = 4.8 m/s

KE_B = 0.5 * 0.2 kg * (4.8 m/s)^2 = 2.304 J

Therefore, the work done by friction can be calculated as the difference in potential energy:
Work = PE_A - KE_B = 3.136 J - 2.304 J = 0.832 J

Therefore, the work done by friction on the package as it slides down the circular track from A to B is 0.832 Joules.

b) To find the coefficient of kinetic friction on the horizontal surface, we can use the work-energy theorem. The work done by friction on the horizontal surface is equal to the change in kinetic energy of the object.

Since the object comes to rest at point C, its final kinetic energy is zero.

The initial kinetic energy of the object can be calculated using the formula:
KE = 0.5 * m * v^2

Where:
m = mass of the object = 0.2 kg
v = velocity at point B = 4.8 m/s

KE_initial = 0.5 * 0.2 kg * (4.8 m/s)^2 = 2.304 J

The work done by friction on the horizontal surface can be calculated as the change in kinetic energy:
Work = KE_initial - KE_final = 2.304 J - 0 J = 2.304 J

The work done by friction is given by:
Work = force of friction * distance

The distance traveled on the horizontal surface is 3 m. So we can rearrange the equation to solve for the force of friction:

force of friction = Work / distance = 2.304 J / 3 m = 0.768 N

The force of friction can be calculated using the formula:
force of friction = coefficient of kinetic friction * normal force

Since the object is on a horizontal surface, the normal force and gravitational force are equal in magnitude but opposite in direction.

Therefore, the coefficient of kinetic friction on the horizontal surface is given by:
coefficient of kinetic friction = force of friction / normal force

To find the normal force, we can use the equation:
normal force = mass * gravitational acceleration

normal force = 0.2 kg * 9.8 m/s^2 = 1.96 N

Therefore, the coefficient of kinetic friction on the horizontal surface is:
coefficient of kinetic friction = 0.768 N / 1.96 N = 0.392

c) To determine the average power produced by friction as the object moves from point B to point C, we need to calculate the work done by friction and the time taken to move the distance.

The work done by friction can be calculated using the formula:
Work = force * distance

Where:
force = force of friction = 0.768 N (calculated in part b)
distance = 3 m

Work = 0.768 N * 3 m = 2.304 J

The time taken to cover a distance of 3 m can be calculated using the formula:
time = distance / velocity

Where:
distance = 3 m
velocity = 4.8 m/s (from point B to C)

time = 3 m / 4.8 m/s = 0.625 s

Power is defined as the rate at which work is done, so the average power produced by friction is given by:
average power = Work / time = 2.304 J / 0.625 s = 3.6864 W

Therefore, the average power produced by friction as the object moves from point B to point C is approximately 3.6864 Watts.

d) The instantaneous power produced by friction at point B can be calculated using the equation:
power = force * velocity

Where:
force = force of friction = 0.768 N (calculated in part b)
velocity = 4.8 m/s (at point B)

power = 0.768 N * 4.8 m/s = 3.6864 W

Therefore, the instantaneous power produced by friction at point B is approximately 3.6864 Watts.

To answer these questions, we'll need to use the laws of motion and the concept of work and energy.

a) To find the work done by friction on the package as it slides down the circular track from A to B, we can use the work-energy theorem. According to the theorem, the work done on an object is equal to the change in its kinetic energy. In this case, the work done by friction will be equal to the change in kinetic energy of the object.

First, let's find the initial and final kinetic energies of the object. The initial speed (at point A) is zero, so the initial kinetic energy is zero. The final speed (at point B) is given as 4.8 m/s. Therefore, the final kinetic energy is given by:

\( KE_{final} = \frac{1}{2} m v_{final}^2 \)

where m is the mass of the object (0.2 kg) and \( v_{final} \) is the final speed (4.8 m/s).

Substituting the values, we get:

\( KE_{final} = \frac{1}{2} \times 0.2 \times (4.8)^2 \)

Solving this equation gives us the final kinetic energy.

The work done by friction is equal to the change in kinetic energy, so we can calculate it by subtracting the initial kinetic energy (zero) from the final kinetic energy.

b) To find the coefficient of kinetic friction on the horizontal surface, we can use the equation for frictional force:

\( f_{friction} = \mu_k \times N \)

where \( f_{friction} \) is the frictional force, \( \mu_k \) is the coefficient of kinetic friction, and N is the normal force.

Since the object is at rest at point C, the frictional force must be equal to the force that would cause it to move (towards the opposite direction). The normal force is equal to the weight of the object, which is given by:

\( N = m \times g \)

where m is the mass of the object (0.2 kg) and g is the acceleration due to gravity (9.8 m/s^2).

Substituting the values into the equation for the frictional force, we can solve for the coefficient of kinetic friction (\( \mu_k \)).

c) To find the average power produced by friction as the object moves from B to C, we can use the equation for power:

\( P = \frac{W}{t} \)

where P is the power, W is the work done, and t is the time taken.

The work done by friction is given by the equation we obtained in part a). We need to find the time taken for the object to move from B to C.

To find the time, we can use the equation of motion for uniform motion:

\( s = ut + \frac{1}{2} a t^2 \)

where s is the distance traveled, u is the initial velocity (4.8 m/s), a is the acceleration (which is zero since the object is at rest at point C), and t is the time taken.

We can rearrange the equation to solve for t, using the given distance traveled (3 m) and the initial velocity.

Then we can substitute the values into the equation for power to find the average power produced by friction.

d) To find the instantaneous power produced by friction at point B, we can use the equation for instantaneous power:

\( P = F \times v \)

where P is the power, F is the force applied, and v is the velocity of the object.

The force applied is the frictional force, which can be calculated using the equation we obtained in part b).

Substituting the values into the equation for power, we can find the instantaneous power produced by friction at point B.