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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?

2 Answers

3 votes

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:


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


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


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


\[ v_f^2 \approx 44.46 \]


\[ v_f \approx √(44.46) \]


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

User Timv
by
8.1k points
3 votes

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.

User Abhishek Sinha
by
7.8k points