5. a woman pushes a 100 gram puck with a force of 2 newtons for 0.5 meters, then lets it slide on its own across the table. friction brings it to a stop at the other end after moving independently for 2 meters.
use energy formulas and the concept of conservation to calculate the following 4 things (include units).
a. the work the woman does on the puck.
b. the speed it leaves her hand.
c. what force does work to stop the puck after it slides.
d. does the normal force from the table do any work? ______ explain why or why not.
6. tell how a machine like an inclined plane may change a force.
7. tell how a machine like an pulley may change a force.

Question

Accepted Answer

### Question 5a: Work the woman does on the puck
# Explanation:
## Step1: Recall work formula
Work $ W $ is calculated as $ W = F \cdot d $, where $ F $ is force and $ d $ is distance in the direction of the force.
## Step2: Substitute values
Given $ F = 2 \, \text{N} $, $ d = 0.5 \, \text{m} $. So $ W = 2 \, \text{N} \times 0.5 \, \text{m} $.
## Step3: Calculate result
$ W = 1 \, \text{Joule (J)} $.
# Answer: $ 1 \, \text{J} $

### Question 5b: Speed the puck leaves her hand
# Explanation:
## Step1: Relate work to kinetic energy
The work done on the puck becomes its kinetic energy ($ KE $), so $ KE = W = 1 \, \text{J} $. Kinetic energy formula: $ KE = \frac{1}{2}mv^2 $.
## Step2: Convert mass to kg
Mass $ m = 100 \, \text{g} = 0.1 \, \text{kg} $.
## Step3: Solve for velocity $ v $
Rearrange $ KE = \frac{1}{2}mv^2 $ to $ v = \sqrt{\frac{2KE}{m}} $. Substitute $ KE = 1 \, \text{J} $, $ m = 0.1 \, \text{kg} $: $ v = \sqrt{\frac{2\times1}{0.1}} = \sqrt{20} \approx 4.47 \, \text{m/s} $.
# Answer: Approximately $ 4.47 \, \text{m/s} $

### Question 5c: Force that stops the puck
# Brief Explanations:
When the puck slides independently, friction acts opposite to its motion, doing negative work to stop it (reduce its kinetic energy to zero). So the force is friction.
# Answer: Friction

### Question 5d: Normal force work
# Brief Explanations:
Work is $ W = Fd\cos\theta $, where $ \theta $ is the angle between force and displacement. Normal force ($ F_N $) is perpendicular to the table (displacement direction is horizontal), so $ \theta = 90^\circ $, $ \cos(90^\circ) = 0 $. Thus, $ W = F_N \cdot d \cdot 0 = 0 $. So normal force does no work.
# Answer: No. Because normal force is perpendicular to displacement, $ W = Fd\cos\theta = Fd\cos(90^\circ) = 0 $.

### Question 6: Inclined plane changing force
# Brief Explanations:
An inclined plane reduces the **magnitude** of the force needed to lift an object (vs. lifting vertically) by increasing the distance over which the force is applied. Work done (force × distance) remains equal (ideally, ignoring friction) for lifting the object to the same height, so a smaller force over a longer distance (the incline length) replaces a larger force over a shorter distance (vertical height).
# Answer: An inclined plane reduces the force needed to lift an object by increasing the distance the force is applied (while keeping work done constant, ideally).

### Question 7: Pulley changing force
# Brief Explanations:
A pulley (especially a block - and - tackle system) changes force by **distributing** the load's weight among multiple rope segments. A movable pulley (or system) reduces the force needed to lift a load: the mechanical advantage (load force / effort force) increases with the number of supporting rope segments, so effort force = load force / number of supporting segments.
# Answer: A pulley (e.g., block - and - tackle) reduces the force needed to lift a load by distributing the load's weight across multiple rope segments (increasing mechanical advantage).

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QUESTION IMAGE

Question

Explanation:

Question 5a: Work the woman does on the puck

Step1: Recall work formula

Step2: Substitute values

Step3: Calculate result

Step1: Relate work to kinetic energy

Step2: Convert mass to kg

Step3: Solve for velocity \( v \)

Answer:

Question 5b: Speed the puck leaves her hand