Problem: A 15N block is being pushed up on a plane inclined at an angle of 20 with the horizontal by a 40N force FA applied parallel to the plane. The block moves 5.0m up the inclined plane. The coefficient of kinetic friction is 0.2. How much work is done by the following forces:
a.) gravitational force
b.) applied force
c.) friction fo
d.) normal force Describe and provide the complete process of your solution with context (not just a line of equations).
Please include the following elements in your solution sheet:
i. Free-Body Diagram with complete labels
ii. Given quantities and desired (unknown) quantities
iii. Complete solution with context/description
iv. Final Answer/s with correct units of measurement (Box your final answer/s for easy identification).
a.) gravitational force
b.) applied force
c.) friction fo
d.) normal force Describe and provide the complete process of your solution with context (not just a line of equations).
Please include the following elements in your solution sheet:
i. Free-Body Diagram with complete labels
ii. Given quantities and desired (unknown) quantities
iii. Complete solution with context/description
iv. Final Answer/s with correct units of measurement (Box your final answer/s for easy identification).
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Answer:
sana maka help
Explanation:
i. Free-Body Diagram:
|\
| \
| \
| \
| \
| \
| \
| \
| \
| \
| θ \
|_________\
N
^
|
---------------
| |
| Block |
| |
---------------
Labels:
- θ: Angle of the inclined plane
- N: Normal force
- mg: Weight of the block (gravitational force)
- FA: Applied force
- f: Force of friction
ii. Given quantities and desired (unknown) quantities:
Given:
- Weight of the block (mg) = 15 N
- Applied force (FA) = 40 N
- Angle of the inclined plane (θ) = 20°
- Distance moved up the inclined plane = 5.0 m
- Coefficient of kinetic friction (μ) = 0.2
Desired:
- Work done by the gravitational force (Wg)
- Work done by the applied force (WA)
- Work done by the force of friction (Wf)
- Work done by the normal force (WN)
iii. Complete solution with context/description:
To calculate the work done by each force, we need to use the formula:
Work (W) = Force (F) * Distance (d) * cos(θ)
a.) Work done by the gravitational force (Wg):
The gravitational force acts vertically downwards. Since the displacement is up the inclined plane, the angle between the force and the displacement is 180°. Therefore, cos(180°) = -1.
Wg = mg * d * cos(180°)
Wg = -15 N * 5.0 m * cos(180°)
Wg = -75 J (Negative sign indicates work done against the force of gravity)
b.) Work done by the applied force (WA):
The applied force is parallel to the displacement, so the angle between the force and the displacement is 0°. Therefore, cos(0°) = 1.
WA = FA * d * cos(0°)
WA = 40 N * 5.0 m * cos(0°)
WA = 200 J
c.) Work done by the force of friction (Wf):
The force of friction acts opposite to the direction of motion, so its angle with the displacement is 180°. Therefore, cos(180°) = -1.
Wf = f * d * cos(180°)
Wf = μ * N * d * cos(180°)
Wf = 0.2 * N * 5.0 m * cos(180°)
Wf = -1 J (Negative sign indicates work done against the force of friction)
d.) Work done by the normal force (WN):
The normal force is perpendicular to the displacement, so the angle between the force and the displacement is 90°. Therefore, cos(90°) = 0.
WN = N * d * cos(90°)
WN = 0 J (No work is done by the normal force)
iv. Final Answer/s with correct units of measurement:
a.) Work done by the gravitational force (Wg) = -75 J
b.) Work done by the applied force (WA) = 200 J
c.) Work done by the force of friction (Wf) = -1 J
d.) Work done by the normal force (WN) = 0 J