A trunk of mass m = 1.100 kg is pushed a distance d = 32.0 cm up an incline with an angle of inclination theta = 26.0 ° by a constant horizontal force P = 150. N. The coefficient of kinetic friction between the trunk and the incline is 0.330.

a) Calculate the work done on the trunk by the applied force P
b) Calculate the work done on the trunk by the frictional force. Be careful – you will need to write down a free-body diagram and calculate the normal force for this part !
c) Calculate the work done on the trunk by the gravitational force

W(P)=P•cosα•d=...

W(fr)= - F(fr) •d= - μ•N•d =
= - μ•( m•g•cosα+P•sinα) •d
W(gr) = - m•g•h = - m•g•d•sin α = ...

a) Okay, let's calculate the work done by the applied force P. The work done by a force is given by the formula W = F * d * cos(theta), where F is the force, d is the displacement, and theta is the angle between the force and displacement vectors.

In this case, the applied force P is horizontal, so the angle theta between P and the displacement is 0 degrees. Plugging in the values, we have W = 150 N * 0.32 m * cos(0°).

But wait, 0° is a special angle with a special value for cos! The cosine of 0° is 1. So we can simplify further to W = 150 N * 0.32 m * 1.

Multiplying these values, we get W = 48 Joules. So the work done on the trunk by the applied force is 48 Joules.

b) Now, let's work out the work done on the trunk by the frictional force. Since the trunk is moving up the incline, the frictional force acts in the opposite direction, so its angle theta is 180 degrees. Applying the same formula, W = F * d * cos(theta), we have W = f * d * cos(180°).

But hold on! The cosine of 180° is -1. So, W = f * d * (-1).

To calculate the frictional force f, we use the formula f = u * N, where u is the coefficient of kinetic friction and N is the normal force. To find N, we need to draw a free-body diagram. But I'm no artist, so you'll have to imagine it.

In our free-body diagram, we have the weight of the trunk pointing downwards and the normal force pointing upwards. Since the trunk is on an incline, we can decompose the weight into perpendicular and parallel components. The perpendicular component is N and the parallel component is m * g * sin(theta), where m is the mass and g is the acceleration due to gravity.

Plugging in the values, we have N = m * g * cos(theta) = 1.1 kg * 9.8 m/s^2 * cos(26°) = 10.069 N.

Now, let's calculate the frictional force f. f = u * N = 0.33 * 10.069 N = 3.32077 N.

Substituting back into our previous equation, W = f * d * (-1) = 3.32077 N * 0.32 m * (-1).

Multiplying these values, we get W = -1.06 Joules (approximately). So, the work done on the trunk by the frictional force is approximately -1.06 Joules.

c) Finally, let's find the work done on the trunk by the gravitational force. The gravitational force is acting vertically downwards, which is perpendicular to the displacement. So, the angle theta between the force and displacement vectors is 90 degrees.

Using the same formula as before, W = F * d * cos(theta), we have W = m * g * d * cos(90°).

But hey, the cosine of 90° is 0! So, W = m * g * d * 0.

Multiplying these values, we get W = 0 Joules. So, the work done on the trunk by the gravitational force is 0 Joules.

And there you have it! The work done by the applied force is 48 Joules, the work done by the frictional force is approximately -1.06 Joules, and the work done by the gravitational force is 0 Joules.

To solve this problem, we need to break it down into different parts and calculate the work done on the trunk by each force. We'll start with part a) and then move on to parts b) and c).

a) Calculate the work done on the trunk by the applied force P:

Work done (W) = Force (F) * Displacement (d) * Cosine of the angle between the force and displacement.

In this case, the angle between the force P and the displacement d is 0 degrees because the force is applied horizontally.

So, W = P * d * cos(0)

Given:
P = 150 N
d = 32.0 cm = 0.32 m

W = 150 * 0.32 * cos(0)
W = 48 J

Therefore, the work done on the trunk by the applied force P is 48 Joules.

b) Calculate the work done on the trunk by the frictional force:
To calculate the work done by the frictional force, we first need to calculate the normal force.

The normal force (N) can be calculated as the gravitational force component perpendicular to the incline, which is given by:

N = mass (m) * gravitational acceleration (g) * cosine(theta)

Given:
m = 1.100 kg
g = 9.8 m/s^2
theta = 26.0 degrees

N = 1.100 * 9.8 * cos(26.0)
N ≈ 9.858 N

Now, using the coefficient of kinetic friction (μ) and the normal force (N), we can calculate the frictional force (f):

f = μ * N

Given:
μ = 0.330

f = 0.330 * 9.858
f ≈ 3.247 N

Now, we can calculate the work done by the frictional force:

W_f = f * d * cos(180 - theta)

Here, the angle between the frictional force and the displacement is 180 - theta because friction opposes motion.

W_f = 3.247 * 0.32 * cos(180 - 26.0)
W_f = 3.247 * 0.32 * cos(154.0)
W_f ≈ -3.051 J

Note that the negative sign indicates that the work done by the frictional force is negative, meaning it is removing energy from the system.

Therefore, the work done on the trunk by the frictional force is approximately -3.051 Joules.

c) Calculate the work done on the trunk by the gravitational force:

The work done by the gravitational force can be calculated as:

W_g = m * g * d * sin(theta)

W_g = 1.100 * 9.8 * 0.32 * sin(26.0)
W_g ≈ 1.809 J

Therefore, the work done on the trunk by the gravitational force is approximately 1.809 Joules.

To calculate the work done on the trunk by different forces, we need to understand the formula for work. Work is defined as the dot product of force and displacement:

Work = Force * Displacement * cos(theta)

a) Calculating the work done on the trunk by the applied force, P:
Given:
Applied force, P = 150 N
Distance, d = 32.0 cm = 0.32 m
Angle of inclination, theta = 26.0 degrees

We know that the applied force is in the horizontal direction, so the angle between the force and displacement is 0 degrees. Therefore, cos(theta) = cos(0) = 1.

Using the formula for work, we have:
Work by applied force, W_applied = P * d * cos(theta)
= 150 N * 0.32 m * cos(26.0)
≈ 118.40 J

Therefore, the work done on the trunk by the applied force, P, is approximately 118.40 Joules.

b) Calculating the work done on the trunk by the frictional force:
To calculate this, we need to find the normal force acting on the trunk. In order to do that, we can draw a free-body diagram for the trunk on the incline:

|__ __ |
/ |
/ |
/________|

The forces acting on the trunk are:
- Gravity (mg) acting vertically downwards
- Normal force (N) acting perpendicular to the incline
- Frictional force (f) acting parallel to the incline

The normal force can be calculated by resolving the weight (mg) into its components. The component acting perpendicular to the incline is equal to the normal force:

N = mg * cos(theta)

Given:
Mass of trunk, m = 1.100 kg
Angle of inclination, theta = 26.0 degrees

Using trigonometry, we have:
N = (1.100 kg * 9.8 m/s^2) * cos(26.0)

The frictional force can be calculated using the equation:
f = coefficient of friction * N

Given:
Coefficient of kinetic friction, mu_k = 0.330

Using this, we can calculate the frictional force:
f = 0.330 * N

Now, let's calculate the work done by the frictional force:
The displacement and the frictional force act in opposite directions, so the angle between them is 180 degrees (or pi radians), meaning cos(180) = -1.

Work by frictional force, W_friction = f * d * cos(180)
= 0.330 * N * d * cos(180)
= 0.330 * (1.100 kg * 9.8 m/s^2 * cos(26.0)) * 0.32 m * cos(180)
≈ -9.15 J

Therefore, the work done on the trunk by the frictional force is approximately -9.15 Joules. The negative sign indicates that the work is done against the direction of the displacement.

c) Calculating the work done on the trunk by the gravitational force:
The work done by the gravitational force is given by:

Work by gravitational force, W_gravity = mg * d * sin(theta)

Using the given values, we can calculate this:
W_gravity = (1.100 kg * 9.8 m/s^2) * 0.32 m * sin(26.0)
≈ 10.96 J

Therefore, the work done on the trunk by the gravitational force is approximately 10.96 Joules.