### Physics Problem on Impulse-Momentum and Braking Force Calculation**Problem Statement:**A 13,129-pound truck enters an emergency exit ramp at an initial speed of 112.4 feet per second (ft/s). It travels for 7.9 seconds before its speed is reduced to 24.5 ft/s. The task is to determine the braking force exerted by the truck, assuming the acceleration is constant. This problem uses the Impulse-Momentum concepts.**Assumptions:**- The angle \( \theta \) of the ramp is 21.4 degrees.**Objective:**- Calculate the braking force (in pounds of force, lbf) to two decimal places.**Diagram Description:**- The diagram illustrates a truck labeled "Rail Lines Cross Country Movers" ascending an inclined ramp.- The initial velocity \( v_0 \) is shown as a green arrow pointing in the direction of the ramp's slope.- The ramp has an indicated angle \( \theta \) with the horizontal plane, specified as 21.4 degrees.**Steps for Calculation:**1. **Extract Given Data:** - Mass of the truck: 13,129 lb - Initial velocity (\( v_0 \)): 112.4 ft/s - Final velocity (\( v_f \)): 24.5 ft/s - Time (\( t \)): 7.9 s - Ramp angle (\( \theta \)): 21.4 degrees2. **Use Impulse-Momentum Theorem:** - Impulse (\( J \)) is the change in momentum (\( \Delta p \)) of the truck. - \( J = \Delta p = m \cdot \Delta v \) - Impulse is also equal to the braking force (\( F \)) times the time (\( t \)): \( J = F \cdot t \)3. **Calculate Change in Velocity:** - \( \Delta v = v_f - v_0 = 24.5 \text{ ft/s} - 112.4 \text{ ft/s} = -87.9 \text{ ft/s} \)4. **Calculate Momentum Change:** - Since \( m \) is given in pounds (force), ensure units are consistent. - Using \( m \) directly (assuming weight and mass conversion factors

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A 13,129-lb truck enters an emergency exit ramp at a speed of 112.4 ft/s. It travels for 7.9 s before its speed is reduced to 24.5 ft/s. Determine the braking force by the truck if the acceleration is constant. (Use Impulse-Momentum concepts.)

A 13,129-lb truck enters an emergency exit ramp at a speed of 112.4 ft/s. It travels for 7.9 s before its speed is reduced to 24.5 ft/s. Determine the braking force by the truck if the acceleration is constant. (Use Impulse-Momentum concepts.)

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

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Concept explainers

Clutches, Brakes, Couplings, and Flywheels

The clutch is described as a machine element used to connect or disconnect two shafts together to transmit torque or power. The clutch works on the principles of friction. The clutch is commonly used in automobiles to transfer power from the driving shaft to the driven shaft.

Question

### Physics Problem on Impulse-Momentum and Braking Force Calculation

**Problem Statement:**
A 13,129-pound truck enters an emergency exit ramp at an initial speed of 112.4 feet per second (ft/s). It travels for 7.9 seconds before its speed is reduced to 24.5 ft/s. The task is to determine the braking force exerted by the truck, assuming the acceleration is constant. This problem uses the Impulse-Momentum concepts.

**Assumptions:**
- The angle \( \theta \) of the ramp is 21.4 degrees.

**Objective:**
- Calculate the braking force (in pounds of force, lbf) to two decimal places.

**Diagram Description:**
- The diagram illustrates a truck labeled "Rail Lines Cross Country Movers" ascending an inclined ramp.
- The initial velocity \( v_0 \) is shown as a green arrow pointing in the direction of the ramp's slope.
- The ramp has an indicated angle \( \theta \) with the horizontal plane, specified as 21.4 degrees.

**Steps for Calculation:**
1. **Extract Given Data:**
- Mass of the truck: 13,129 lb
- Initial velocity (\( v_0 \)): 112.4 ft/s
- Final velocity (\( v_f \)): 24.5 ft/s
- Time (\( t \)): 7.9 s
- Ramp angle (\( \theta \)): 21.4 degrees

2. **Use Impulse-Momentum Theorem:**
- Impulse (\( J \)) is the change in momentum (\( \Delta p \)) of the truck.
- \( J = \Delta p = m \cdot \Delta v \)
- Impulse is also equal to the braking force (\( F \)) times the time (\( t \)): \( J = F \cdot t \)

3. **Calculate Change in Velocity:**
- \( \Delta v = v_f - v_0 = 24.5 \text{ ft/s} - 112.4 \text{ ft/s} = -87.9 \text{ ft/s} \)

4. **Calculate Momentum Change:**
- Since \( m \) is given in pounds (force), ensure units are consistent.
- Using \( m \) directly (assuming weight and mass conversion factors

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