A Porsche 944 Turbo has a rated engine power of 217 hp. 30% of the power is lost in the engine and the drive train, and 70% reaches the wheels. The total mass of the car and driver is 1480 kg, and two-thirds of the weight is over the drive wheels. (a) What is the maximum acceleration of the Porsche on a concrete surface where μₛ = 1.00 ? Hint: What force pushes the car forward?

Question

A Porsche 944 Turbo has a rated engine power of 217 hp. 30% of the power is lost in the engine and the drive train, and 70% reaches the wheels. The total mass of the car and driver is 1480 kg, and two-thirds of the weight is over the drive wheels. (a) What is the maximum acceleration of the Porsche on a concrete surface where μₛ = 1.00 ? Hint: What force pushes the car forward?

Accepted Answer

Hey, everyone. So this is a pretty straightforward static friction problem. Let's see what they're asking us. A loaded heavy construction vehicle has a mass of kg. The power output of the engine is 420 hp. The engine and transmission dissipate 50% of the power output delivering 85% to the wheels. Suppose that 80% of the vehicle's weight is carried by the drive wheels. And we are asked to determine the greatest acceleration of the vehicle on a Tarmac Road where the coefficient of static friction is 1.2. Our multiple choice answers here are a one point sorry, a 11.8 m per second squared. B 9.41 m per second squared. C 1.2 m per second squared or D 0.111, sorry D 0.122 m per second squared. Ok. So the first thing we're going to do in this problem is draw our free body diagram. We'll use the construction vehicles, drive wheels as our, as our point. So we have our static friction force acting in the same direction as the attempted motion we have, so that's the positive X direction, we have our normal force acting in the positive Y direction and then our weight acting in the negative Y direction. We can recall from Newton's second law that the sum of the forces is equal to mass multiplied by acceleration. So we'll write that equation for both the forces in the X direction and the forces in the Y direction, some of the forces in the Y direction is also equal to mass times acceleration in the Y direction. So that's it why component of the acceleration? So the sum of the forces in the direction, the only force in the direction is our static friction force. So we have our static friction force is equal to mass multiplied by acceleration. And then in the Y direction, we have our normal force which is positive minus our weight is equal to Mass times acceleration or acceleration in the Y. direction. However, is zero. So our, so this equation becomes normal force minus weight equals zero. Our static friction force we can recall is given by us multiplied by the normal force and that is equal again to mass times acceleration. And we will use our why component forces to solve for our normal force. So normal force is equal to the weight Which we can recall is equal to mass times gravity. However, we need to recall here or we need to recognize here that only 80% of the vehicle's weight is carried by the drive wheels. So our normal force becomes 80% or 0. multiplied by mass multiplied by gravity. So in solving for the acceleration, that's what the problem is asking us for. That greatest acceleration will rearrange this equation to be acceleration is equal to us N Over M and then we'll plug in our normal force. So that will be us multiplied by 0.8, multiplied by mass multiplied by gravity over mass. And then those masses cancel. And so then from there, we can plug in our values that we have. We know our static friction coefficient is 1.2 multiplied by 0.8 multiplied by gravity, which is a constant. We can recall 9.8 m per second squared, Plug that into our calculators and we get 9.41 meters per second squared. And that is the answer to this problem. And that aligns with answer choice B. So that's all we have for this one. We'll see you in the next video.

Verified Solution

Key Concepts

Power and Efficiency

Friction and Traction

Newton's Second Law of Motion