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That the ball attains:force of 3000 is applied to 1500 Kg car at rest: (a) What its acceleration? (b) What will itsvelocity be 5 $ later?...

Question

That the ball attains:force of 3000 is applied to 1500 Kg car at rest: (a) What its acceleration? (b) What will itsvelocity be 5 $ later?

that the ball attains: force of 3000 is applied to 1500 Kg car at rest: (a) What its acceleration? (b) What will its velocity be 5 $ later?



Answers

(a) What force is required to stop a 1500-kg car in a distance of 0.20 m if it is initially moving at 2.2 m/s? (b) What if the car is moving at 4.5 m/s?

In this problem, we have a car with a mass off 1000 500 kg which is traveling for a distance as zero point 20 m before being stopped. The initial velocity off the car is 2.2 m per second and the final velocity is obviously zero because the car is being stopped. The question asks us to evaluate the force needed for the car to stop. Now we start from the equation off the impulse that links the impulse to these force. In fact, the impulse is equal to the average force times Delta T, which is the time interval for the car to stop. And we know that this is actually equal to the change in momentum because that's the finish inal impulse. So we can rearrange this equation to get the force this average force that we need to calculate, which is equal to Delta P changing momentum over Delta T time interval and that is the equation that we will use to determine the force. So we need to do this for part a off the problem where the velocity, the initial velocity is 2.2 m per second and then we will do it in part B for a different value off the initial velocity. Let's start with the evaluation off. The change in momentum Delta P, which is defined as final momentum P f minus initial momentum P I Yeah, final momentum is M V. Where V is the final speed on the initial momentum is, um, you where you is the initial speed. So this can be written as, um times the minus you and this will be put in the numbers in 1500. The mass times the final velocity is zero. An initial velocity is 2.2. So this is a negative change in momentum, which just stands for its direction, being opposite to that of motion off the car. And the magnitude is 3300 kg meters per second. So the negative sign here just states that the momentum, the change in momentum is in the opposite direction to the motion off the car. Let's assume the car is moving to the right. That's positive direction. The changing momentum will be to the left, and that is what the negative sign stands for. We are then going to evaluate the time interval. Now to evaluate the time interval, we use one of the four equation of uniformly accelerated motion. The equation that links as the distance struggled to the two velocities u and V and the time Delta t. So we re arrange this for the time we get Delta T is equal to two s. Where s is our distance? They were 0.20 divided by U plus the and this is equal to putting the numbers in two times 0.2, which is the distance divided by 2.2 plus in Europe because he is zero and that gives us zero point 18 seconds. So now we have the change in momentum and a time and we can calculate the force. The force is equal to Delta P over Delta T as set by this equation over here. And so we have 3300 over zero point 18. Now, I am considering the magnitude off the change in momentum here. That's why I didn't write the minus sign. And this is about 1.8 times 10 to the power of four Newtons. And this is the value off the force required to stop the car Now obviously the direction off the force will be to the left. In fact, if we put the negative sign for the change in momentum, we get a negative sign for the force, which means that the force is to the left. So this is the magnitude off the force 1.8 and sent to the power for Newton and its direction would obviously be to the left. We are now going to evaluate the force for the second part off the question, which is exactly the same setting. But this time the car was initially moving at 4.5 m per second. So it's the initial speed off the car that changes. It is now 4.5. Everything else stays the same, including the equations that we need to use. So we start from the change in momentum Delta P, which we can straight away right as m the minus you. As we know, putting the numbers in that will be 1500. The mask the car times zero minus 4.5 because that's the new value off the velocity, and that will be 6000 700 and 50 kg meters per second. That's the new change in momentum. Now we get also a new value for the time interval needed to stop the car that is equal to two s over u plus the two times 0.2 because the distance has not changed. And this time we have 4.5 plus zero, and that gives us about 0.9 seconds and as the new value off the time. So finally, we can write down the equation for the force again, which is Delta P over Delta T. So we have the new changing momentum 6750 divided by the new time intervals 0.9 And that gives us an increase value off the force, which is 7.5 times 10 to the power of form Newton's. And this force is also to the left. In fact, here again, if we consider the sign of the change in momentum, it would actually be negative. Here are just considered the magnitude. The force is obviously increased because the initial velocity off the car you was bigger and for a bigger initial velocity, it does make sense that the force needed to stop. The car would need to be bigger as well

This question covers the concept of the work energy forum and which states that the networked and is the camera to the changing kind of technology. And the second concept is the work done by a constraint fools and that is equivalent to the magnitude of the foods into their displacement, into the coastline of angle between the force and the displacement. So for part A the work done by the mechanic equals the final kind of technology that is half of mass into the final velocity squared minus half of mass, into the initial velocity squared, since initially the courage address. So we can neglect the storm. So the final velocity of the final speed equals two into the work done upon the mass and the square root. Now we can substitute the value. So the final speed equals two into the work done. And the work done is 5000 jewels upon the masses and the masses. Ah 2500 kg. So the final speed of the car equals two meters for a second. Now, part B, the network done equals the fools Into the displacement, into the course of the angle between them and the angle between them is 0°.. So we can write the the force. The force equals the work done upon the displacement. Now we can substitute the value to find the force. And the force is, The work done is 5000 jewels, And the displacement is 25 meters, so the foods exerted equals 200 newtons. Yeah.

In part, a acceleration is the change in velocity over the amount of time. The change in velocity is, um, 0 to 100 kilometers per hour and the change in time is four seconds. Now there are 3000 600 Excited when I change that 3600 seconds in an hour, hours will cancel out and there are 1000 meters in a kilometer. So I'm going to put this into a calculator. 100 100,000 divided by four and 3600 6.9, um, meters per second. Be if it stopped in five seconds with the same thing. What would have its acceleration be? Well, its acceleration would be negative 100 kilometers per hour over five seconds. So we have to divide by the same 3600 and multiply by the same 1000. And so put that into a calculator. You'd be negative 5.6 meters per second. See? Yeah. If the mass of the car is 1200 kilograms than what is the Net force on the car? No, the force is the mass times acceleration. So when it's stopping which I guess this is what it's asking when it's when it's stopping. Then I have to take the acceleration, which is negative 5.6 times 1200. And that gives me negative 6700 Newtons de what applies the push that accelerates the car? Um, that would be the engine through gears and pistons and that sort of stuff that provides ah and axles that rotates the wheels that pushes against the friction on the road that makes the car go forward.

Number 88. We will use the concept off impulse here from the definition off the impulse we know that force into the time period is equal to the change in momentum which is Marcin toe final last a minus initial lasting. So now, subjecting the values course time. Peter this. 10 seconds Ma says 1500 kg and finally lost his 30 m per second initial lost zero. Solving this for force, we will get forces equal toe 45 100 Newton.


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