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A car of mass 1000 kg moving with velocity 72 km per hour takes 4 second to stop after the bricks are applied the force exerted by the brakes is?
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A car of mass 1000 kg moving with velocity 72 km per hour takes 4 seco...
To find the force exerted by the brakes on the car, we can follow these steps:
Convert Velocity
- The initial velocity (u) of the car is given as 72 km/h.
- Convert this to meters per second (m/s):
- \( u = 72 \text{ km/h} \times \frac{1000 \text{ m}}{3600 \text{ s}} = 20 \text{ m/s} \)
Calculate the Deceleration
- The final velocity (v) after coming to a stop is 0 m/s.
- The time (t) taken to stop is 4 seconds.
- Using the formula for acceleration (or deceleration in this case):
- \( a = \frac{v - u}{t} \)
- \( a = \frac{0 \text{ m/s} - 20 \text{ m/s}}{4 \text{ s}} = -5 \text{ m/s}^2 \)
Calculate the Force
- To find the force exerted by the brakes, we use Newton's second law:
- \( F = m \cdot a \)
- Where \( m \) is the mass of the car (1000 kg) and \( a \) is the deceleration (-5 m/s²).
- Therefore:
- \( F = 1000 \text{ kg} \cdot (-5 \text{ m/s}^2) = -5000 \text{ N} \)
Conclusion
- The negative sign indicates that the force exerted by the brakes is in the opposite direction of the car’s motion.
- Thus, the magnitude of the force exerted by the brakes is **5000 N**.
Convert Velocity
- The initial velocity (u) of the car is given as 72 km/h.
- Convert this to meters per second (m/s):
- \( u = 72 \text{ km/h} \times \frac{1000 \text{ m}}{3600 \text{ s}} = 20 \text{ m/s} \)
Calculate the Deceleration
- The final velocity (v) after coming to a stop is 0 m/s.
- The time (t) taken to stop is 4 seconds.
- Using the formula for acceleration (or deceleration in this case):
- \( a = \frac{v - u}{t} \)
- \( a = \frac{0 \text{ m/s} - 20 \text{ m/s}}{4 \text{ s}} = -5 \text{ m/s}^2 \)
Calculate the Force
- To find the force exerted by the brakes, we use Newton's second law:
- \( F = m \cdot a \)
- Where \( m \) is the mass of the car (1000 kg) and \( a \) is the deceleration (-5 m/s²).
- Therefore:
- \( F = 1000 \text{ kg} \cdot (-5 \text{ m/s}^2) = -5000 \text{ N} \)
Conclusion
- The negative sign indicates that the force exerted by the brakes is in the opposite direction of the car’s motion.
- Thus, the magnitude of the force exerted by the brakes is **5000 N**.
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A car of mass 1000 kg moving with velocity 72 km per hour takes 4 second to stop after the bricks are applied the force exerted by the brakes is?
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A car of mass 1000 kg moving with velocity 72 km per hour takes 4 second to stop after the bricks are applied the force exerted by the brakes is? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about A car of mass 1000 kg moving with velocity 72 km per hour takes 4 second to stop after the bricks are applied the force exerted by the brakes is? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A car of mass 1000 kg moving with velocity 72 km per hour takes 4 second to stop after the bricks are applied the force exerted by the brakes is?.
A car of mass 1000 kg moving with velocity 72 km per hour takes 4 second to stop after the bricks are applied the force exerted by the brakes is? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about A car of mass 1000 kg moving with velocity 72 km per hour takes 4 second to stop after the bricks are applied the force exerted by the brakes is? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A car of mass 1000 kg moving with velocity 72 km per hour takes 4 second to stop after the bricks are applied the force exerted by the brakes is?.
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