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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
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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### 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
Transcribed Image Text:### 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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