A block of weight = 35.0 N sits on a frictionless inclined plane, which makes an angle theta = 27.0 with respect to the horizontal. A force of magnitude = 15.9 N, applied parallel to the incline, is just sufficient to pull the block up the plane at constant speed. What is Wg, the work done on the block by the force of gravity w as the block moves a distance L= 3.20 m up the incline? What is Wf, the work done on the block by the applied force as the block moves a distance L= 3.20 m up the incline?
To find Wg, the work done on the block by the force of gravity as the block moves up the incline, we can use the formula:
Wg = m * g * d * cos(theta)
Where:
m = mass of the block
g = acceleration due to gravity
d = distance moved up the incline
theta = angle of the incline
First, let's find the mass of the block using the weight given:
Weight = mass * g
35.0 N = m * 9.8 m/s^2
Solving for m, we get:
m = 35.0 N / 9.8 m/s^2
m ≈ 3.57 kg
Now we can substitute the values into the formula for Wg:
Wg = 3.57 kg * 9.8 m/s^2 * 3.20 m * cos(27.0°)
Using a calculator, we can find that cos(27.0°) ≈ 0.890
Wg ≈ 100.73 J
Therefore, the work done on the block by the force of gravity, Wg, is approximately 100.73 J.
To find Wf, the work done on the block by the applied force as the block moves up the incline, we can use the formula:
Wf = F * d * cos(theta)
Where:
F = applied force
d = distance moved up the incline
theta = angle of the incline
Substituting the given values into the formula for Wf:
Wf = 15.9 N * 3.20 m * cos(27.0°)
Using a calculator, we can find that cos(27.0°) ≈ 0.890
Wf ≈ 45.12 J
Therefore, the work done on the block by the applied force, Wf, is approximately 45.12 J.
To find the work done on the block by the force of gravity (Wg) as it moves up the incline, we need to calculate the component of the weight (force of gravity) parallel to the incline.
The weight (force of gravity) can be divided into two components: one perpendicular to the incline (W⊥) and one parallel to the incline (W//). The component parallel to the incline is what contributes to the work done.
First, let's find the component of the weight parallel to the incline (W//). We can use trigonometry to calculate it:
W// = W * sin(θ)
W// = 35.0 N * sin(27.0°)
Now, let's find the work done by the force of gravity (Wg) as the block moves a distance L up the incline:
Wg = W// * L
Substituting the values:
Wg = (35.0 N * sin(27.0°)) * 3.20 m
Calculate the above expression to find the value of Wg.
To find the work done on the block by the applied force (Wf) as it moves up the incline, we can use the formula:
Wf = F * L
Here, F is the magnitude of the applied force, which is given as 15.9 N, and L is the distance the block moves up the incline, which is given as 3.20 m.
Substituting the given values:
Wf = 15.9 N * 3.20 m
Calculate the above expression to find the value of Wf.
Please perform the calculations to find the values of Wg and Wf.