Math Problem Statement

A 135 lb golfer is pushing their 23 lb golf bag up a 20 ∘ hill. The tractive force between the golfer and the ground is 92 lb. Assume the golfer pushes on the bag parallel to the ground and the wheels of the golf bag roll freely. Determine the acceleration of the golf bag.

Solution

To determine the acceleration of the golf bag, we need to apply Newton's second law of motion:

Fnet=mbagaF_{\text{net}} = m_{\text{bag}} a

Where:

  • FnetF_{\text{net}} is the net force acting on the golf bag,
  • mbagm_{\text{bag}} is the mass of the golf bag, and
  • aa is the acceleration of the bag.

Step 1: Breakdown of forces

The forces acting on the golf bag include:

  • The pushing force by the golfer: The golfer exerts a force of 92 lb (since tractive force represents the total horizontal force they can apply) parallel to the ground.
  • The gravitational force on the bag: This is the weight of the golf bag acting vertically downwards.
  • The component of the gravitational force pulling the bag back down the incline: This component acts along the incline and can be calculated as:

Fgravity, parallel=WbagsinθF_{\text{gravity, parallel}} = W_{\text{bag}} \sin \theta

Where Wbag=23lbW_{\text{bag}} = 23 \, \text{lb} (the weight of the golf bag) and θ=20\theta = 20^\circ (the incline angle).

Step 2: Calculate the parallel component of gravity

Using the equation for the parallel component of the gravitational force:

Fgravity, parallel=23lb×sin(20)F_{\text{gravity, parallel}} = 23 \, \text{lb} \times \sin(20^\circ)

Let's calculate this value:

Fgravity, parallel23×0.3420=7.866lbF_{\text{gravity, parallel}} \approx 23 \times 0.3420 = 7.866 \, \text{lb}

Step 3: Determine the net force acting on the bag

The net force FnetF_{\text{net}} is the pushing force by the golfer minus the gravitational component pulling the bag down the incline:

Fnet=92lb7.866lb=84.134lbF_{\text{net}} = 92 \, \text{lb} - 7.866 \, \text{lb} = 84.134 \, \text{lb}

Step 4: Convert weight to mass

We need to convert the weight of the golf bag to its mass for Newton's second law. In the imperial system, the conversion is:

mbag=Wbagg=23lb32.2ft/s20.714slugsm_{\text{bag}} = \frac{W_{\text{bag}}}{g} = \frac{23 \, \text{lb}}{32.2 \, \text{ft/s}^2} \approx 0.714 \, \text{slugs}

Step 5: Apply Newton's second law

Now that we have the net force and the mass of the bag, we can calculate the acceleration:

Fnet=mbagaF_{\text{net}} = m_{\text{bag}} a

a=Fnetmbag=84.134lb0.714slugs117.8ft/s2a = \frac{F_{\text{net}}}{m_{\text{bag}}} = \frac{84.134 \, \text{lb}}{0.714 \, \text{slugs}} \approx 117.8 \, \text{ft/s}^2

Final Answer:

The acceleration of the golf bag is approximately 117.8 ft/s².

Would you like any additional details or further clarification?

Here are 5 related questions for further exploration:

  1. What is the significance of converting pounds to slugs in physics problems?
  2. How would the problem change if the hill incline increased to 30 degrees?
  3. What is the role of friction in this scenario, and why is it neglected here?
  4. How does the tractive force affect the acceleration when different surfaces are involved?
  5. What would happen if the golfer applied a force not parallel to the ground?

Tip: Always remember to convert weight to mass when applying Newton's second law in the imperial system!

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Math Problem Analysis

Mathematical Concepts

Newton's Second Law
Forces on an Incline
Gravitational Components

Formulas

F_net = m_bag * a
F_gravity_parallel = W_bag * sin(θ)
m_bag = W_bag / g

Theorems

Newton's Second Law of Motion

Suitable Grade Level

Grades 11-12