A block of weight w = 35.0 N sits on a frictionless inclined plane, which makes an angle theta = 27.0 degrees with respect to the horizontal. A force of magnitude F = 15.9 N, applied parallel to the incline, is just sufficient to pull the block up the plane at constant speed. What is W_g, the work done on the block by the force of gravity w_vec as the block moves a distance L = 3.20 m up the incline?

Well, isn't this a "weighty" problem!

To find the work done by the force of gravity, we need to calculate the component of the weight vector (Wg) that acts along the direction of the displacement (L) up the incline.

Let's break it down step by step. We know the weight (w) of the block is 35.0 N, and it makes an angle (theta) of 27.0 degrees with the horizontal.

The component of weight acting along the incline can be found using the formula: Wg = mg * sin(theta), where m is the mass of the block and g is the acceleration due to gravity (approximately 9.8 m/s^2).

Since we have the weight (w), we can use the equation w = mg to find the mass (m).

So, m = w / g, which is 35.0 N / 9.8 m/s^2.

Now, we can find the work done by the force of gravity using the formula: Wg = m * g * sin(theta) * L.

Plugging in the values, we get:

Wg = (35.0 N / 9.8 m/s^2) * 9.8 m/s^2 * sin(27.0 degrees) * 3.20 m.

Simplifying this expression gives us the work done by the force of gravity, Wg.

To find the work done on the block by the force of gravity W_g, we can calculate the gravitational potential energy change of the block as it moves a distance L up the incline.

The gravitational potential energy change is given by the formula:

ΔPE = m * g * Δh

where ΔPE is the change in potential energy, m is the mass of the block, g is the acceleration due to gravity, and Δh is the change in height.

We can also express the change in height Δh in terms of the distance along the incline L and the angle theta.

Δh = L * sin(theta)

Given that the weight of the block is 35.0 N, we can calculate the mass m using the formula:

w = m * g

m = w / g

Substituting the given values:

m = 35.0 N / 9.8 m/s^2 ≈ 3.57 kg

Now we can calculate Δh:

Δh = L * sin(theta)

Δh = 3.20 m * sin(27.0°) ≈ 1.49 m

Finally, we can calculate the work done on the block by the force of gravity W_g:

W_g = ΔPE

W_g = m * g * Δh

Substituting the calculated values:

W_g ≈ 3.57 kg * 9.8 m/s^2 * 1.49 m

W_g ≈ 52.84 J

Therefore, the work done on the block by the force of gravity w_vec as it moves a distance L = 3.20 m up the incline is approximately 52.84 J.

To find the work done on the block by the force of gravity, we can use the following formula:

W_g = w_vec * d * cos(theta)

where:
- W_g is the work done by the force of gravity
- w_vec is the weight of the block (35.0 N in this case)
- d is the distance the block moves up the incline (3.20 m in this case)
- theta is the angle between the incline and the horizontal (27.0 degrees in this case)

In order to calculate the work done by the force of gravity, we need to find the vertical component of the weight, w_vec_y, which is given by:

w_vec_y = w * sin(theta)

where:
- w is the weight of the block
- sin(theta) is the sine of the angle between the incline and the horizontal

Let's calculate w_vec_y:

Calculate the vertical rise and apply conservation of energy. Work done = poetential energy increase.

We will gladly critique your work