An 8.00kg package in a mail-sorting room slides 2.00 m down a chute that is inclined at 53.0 degrees below the horizontal. The coefficient of kinetic friction between the package and the chute's surface is 0.400.Calculate the work done on the package by friction. Calculate the work done on the package by gravity.

The work done by gravity is

M g * (decrease in elevation)
= M g 2.00 sin 53

The work done by friction is
-(friction foirce)(distance moved)
-M g cos 53 (0.4)* 2.00
It is negative because work is done against friction

To calculate the work done on the package by friction, we need to find the force of friction first. We can use the formula:

Frictional Force = u * Normal Force,

where u is the coefficient of kinetic friction and the Normal Force is the force perpendicular to the surface, which is equal to the weight of the package in this case.

1. Calculate the Normal Force:

Normal Force = mass * gravity,

where gravity is the acceleration due to gravity (9.8 m/s^2).

Normal Force = 8.00 kg * 9.8 m/s^2 = 78.4 N.

2. Calculate the Frictional Force:

Frictional Force = 0.400 * 78.4 N = 31.4 N.

3. Calculate the work done by friction:

Work done by friction = Frictional Force * distance,

distance = 2.00 m (given).

Work done by friction = 31.4 N * 2.00 m = <<31.4*2=62.8>>62.8 J.

Now, let's calculate the work done on the package by gravity:

Work done by gravity = Force of gravity * distance,

Force of gravity = mass * gravity,

distance = 2.00 m (given).

Work done by gravity = 8.00 kg * 9.8 m/s^2 * 2.00 m = 156.8 J.

Therefore, the work done on the package by friction is 62.8 J, and the work done on the package by gravity is 156.8 J.

To calculate the work done on the package by friction, you need to determine the force of friction acting on the package and multiply it by the distance it travels down the chute.

First, calculate the normal force acting on the package. Since the package is on an inclined surface, the normal force (N) is equal to the perpendicular component of the weight of the package, which can be calculated as:

N = mg cos(theta)

where m is the mass of the package (8.00 kg), g is the acceleration due to gravity (9.8 m/s^2), and theta is the angle of the inclined surface (53.0 degrees).

N = (8.00 kg)(9.8 m/s^2)cos(53.0 degrees)
N ≈ 60.63 N

Next, calculate the force of friction (f) using the equation:

f = μN

where μ is the coefficient of kinetic friction between the package and the chute's surface (0.400).

f = (0.400)(60.63 N)
f ≈ 24.25 N

Now, calculate the work done by friction (W_friction) using the equation:

W_friction = f * d

where d is the distance traveled by the package down the chute (2.00 m).

W_friction = (24.25 N)(2.00 m)
W_friction ≈ 48.50 J

Therefore, the work done on the package by friction is approximately 48.50 J.

To calculate the work done on the package by gravity, you need to determine the component of the weight force parallel to the motion of the package and multiply it by the distance it travels down the chute.

The component of the weight force parallel to the motion of the package can be calculated as:

F_parallel = mg sin(theta)

F_parallel = (8.00 kg)(9.8 m/s^2)sin(53.0 degrees)
F_parallel ≈ 63.81 N

Next, calculate the work done by gravity (W_gravity) using the equation:

W_gravity = F_parallel * d

W_gravity = (63.81 N)(2.00 m)
W_gravity ≈ 127.62 J

Therefore, the work done on the package by gravity is approximately 127.62 J.