A 1000-kg car traveling with a speed of 35 m/s skids to a stop in 25m. Determine the average force applied to the breaks. If it took 3.2s to stop, who much power was generated by the breaks?

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Question:

A 1000-kg car traveling with a speed of 35 m/s skids to a stop in 25m. Determine the average force applied to the breaks. If it took 3.2s to stop, who much power was generated by the breaks?

Answer:1. 24500 N. 2. 191406.25 watt Explanation:From the question given above, the following data were obtained:Mass (m) of car = 1000 KgInitial velocity (u) = 35 m/sDistance (s) = 25 mFinal velocity (v) = 0 m/sTime (t) = 3.2 sForce (F) =? Power (P) =? Next, we shall determine the acceleration of the car. This can be obtained as follow:Initial velocity (u) = 35 m/sDistance (s) = 25 mFinal velocity (v) = 0 m/sAcceleration (a) =? vÂ² = uÂ² + 2as0 = 35Â² + (2 Ã— a Ã— 25) 0 = 1225 + 50aCollect like terms 0 â€“ 1225 = 50aâ€“ 1225 = 50aDivide both side by 50a = â€“ 1225 / 50a = â€“24.5 m/sÂ²1. Determination of the force. Mass (m) of car = 1000 KgAcceleration (a) = â€“24.5 m/sÂ² Force (F) =? F = m Ã— aF = 1000 Ã— â€“24.5F = â€“24500 NThe negative sign indicate that the force is in opposite direction to the motion of the car. 2. Determination of the power Force (F) = 24500 NDistance (s) = 25 mTime (t) = 3.2 sPower (P) =? Next, we shall determine the energy. Energy = force Ã— distanceEnergy = 24500 Ã— 25Energy = 612500 JFinally, we shall determine the power. This is illustrated below:Power = Energy /time Power = 612500 /3.2Power = 191406.25 watt

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