Question

You pull a 125kg trunk along the floor with an applied force of 637N at an angle of 24.3. The trunk was initially at rest. You pull it for a distance of 12.7m. The coefficient of kinetic friction is 0.345.

How much net work was done?
What was the speed of the trunk after being pulled 12.7m?

Answer 1

1. The net work done is approximately
`\( 2778.49 \, \text{J} \)`.

2. The final speed of the trunk after being pulled for 12.7 m is approximately
`\( 6.67 \, \text{m/s} \)`.

To find the net work done and the final speed of the trunk, let's break down the problem:

1. Net Work Done:

Net work is the work done by the applied force minus the work done against friction.

Work done by the applied force:

`\[ W_{\text{applied}} = F_{\text{applied}} * d * \cos(\theta) \]`

Where:

-
`\( F_{\text{applied}} = 637 \, \text{N} \)`

-
`\( d = 12.7 \, \text{m} \)`

-
`\( \theta = 24.3^\circ \)`

Work done against friction:

`\[ W_{\text{friction}} = \mu_k * N * d \]`

Where:

-
`\( \mu_k = 0.345 \)` (coefficient of kinetic friction)

- N is the normal force (equal to the weight of the trunk, N = mg )

-
`\( d = 12.7 \, \text{m} \)`

Net work done:

`\[ \text{Net Work} = W_{\text{applied}} - W_{\text{friction}} \]`

2. Final Speed of the Trunk:

The net work done is equal to the change in kinetic energy of the trunk. Using the work-energy principle:

`\[ \text{Net Work} = \Delta KE = (1)/(2)mv_f^2 - (1)/(2)mv_i^2 \]`

Where:

-
`\( m = 125 \, \text{kg} \)` (mass of the trunk)

-
`\( v_i = 0 \, \text{m/s} \)` (initial speed, the trunk is initially at rest)

-
`\( v_f \)` is the final speed of the trunk

Let's calculate these:

1. Net Work Done:

`\[ W_{\text{applied}} = 637 * 12.7 * \cos(24.3^\circ) \]`

`\[ W_{\text{applied}} \approx 8106.94 \, \text{J} \]`

Normal force
`\( N = mg \)`

`\[ N = 125 * 9.81 \]`

`\[ N \approx 1226.25 \, \text{N} \]`

`\[ W_{\text{friction}} = 0.345 * 1226.25 * 12.7 \]`

`\[ W_{\text{friction}} \approx 5328.45 \, \text{J} \]`

Net Work:

`\[ \text{Net Work} = 8106.94 - 5328.45 \]`

`\[ \text{Net Work} \approx 2778.49 \, \text{J} \]`

2. Final Speed of the Trunk:

Using the work-energy principle:

`\[ \text{Net Work} = (1)/(2)mv_f^2 - (1)/(2)mv_i^2 \]`

`\[ 2778.49 = (1)/(2) * 125 * v_f^2 - 0 \]`

`\[ v_f^2 = (2 * 2778.49)/(125) \]`

`\[ v_f^2 \approx 44.46 \]`

`\[ v_f \approx √(44.46) \]`

`\[ v_f \approx 6.67 \, \text{m/s} \]`

Answer 2

The net work done on the trunk was approximately 2217.35 J. The speed of the trunk after being pulled 12.7 meters was approximately 5.7 m/s.

How to find speed?

Calculating net work and final speed

Part 1: net work done

Convert angle to radians:

`angle_{radians` = 24.3° × (π / 180°)

≈ 0.424 radians

Calculate normal force:

`normal_{force` = mass × gravity

`normal_{force` = 125 kg × 9.81 m/s²

`normal_{force` = 1226.25 N

Calculate friction force:

`friction_(force) = coefficient_(of)_(friction) * normal_{force`

`friction_{force` = 0.345 × 1226.25 N

`friction_{force` = 420.2 N

Calculate force component in the direction of motion:

`force_x = applied_(force) * cos(angle_(radians))`

`force_x` = 637 N × cos(0.424 radians)

`force_x` = 594.7 N

Calculate net force:

`net_(force) = force_x - friction_{force`

`net_{force` = 594.7 N - 420.2 N

`net_{force` = 174.5 N

Calculate net work:

`work_(done) = net_(force) * distance`

`work_{done` = 174.5 N × 12.7 m

`work_{done` = 2217.35 J

Therefore, the net work done on the trunk was approximately 2217.35 J.

Part 2: Final Speed

Initial kinetic energy:

`initial_(kinetic)_{energy` = 0 J (since the trunk was initially at rest)

Work-energy principle:

`work_{done` = change in kinetic energy

2217.35 J =
`final_(kinetic)_{energy` - 0 J

Final kinetic energy:

`final_(kinetic)_{energy` = 2217.35 J

Final velocity:

`final_{velocity` = √(2 ×
`final_(kinetic)_{energy` / mass)

`final_{velocity` = √(2 × 2217.35 J / 125 kg)

`final_{velocity` = 5.7 m/s

Therefore, the speed of the trunk after being pulled 12.7 meters was approximately 5.7 m/s.