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$\bullet$$\bullet$ A pickup truck is carrying a toolbox, but the rear gate ofthe truck is missing, so the box will slide out if it is set moving.The coefficients of kinetic and static friction between the boxand the bed of the truck are 0.355 and 0.650 , respectively.Starting from rest, what is the shortest time this truck couldaccelerate uniformly to 30.0 $\mathrm{m} / \mathrm{s}(\approx 60 \mathrm{mph})$ without causingthe box to slide. Include a free-body diagram of the toolbox aspart of your solution. (Hint: First use Newton's second law tofind the maximum acceleration that static friction can give thebox, and then solve for the time required to reach 30.0 $\mathrm{m} / \mathrm{s} .$ )

4.71 s

Physics 101 Mechanics

Chapter 5

Applications of Newton's Law

Motion Along a Straight Line

Motion in 2d or 3d

Newton's Laws of Motion

Applying Newton's Laws

Rutgers, The State University of New Jersey

Hope College

University of Winnipeg

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pressure, 79 states that a pickup truck is carrying toolbox, but the rear gate of the truck is missing, so the box was slide out for just set movie the coefficients of Connecticut. Static friction between the bed and the truck. Our 0.355 point 65 respectively, starting from rest. What is the shortest time? This truck could accelerate uniformly to 30 meters per second without causing the box the slide and quit a free body diagram of the toolboxes partner solution. And as a hint, they say that first Years, Newton's second law, to find the maximum acceleration that's static friction can give the box, and it's all for that time required to reach 30 meters per second. So the hint is we want to find acceleration and then use that to solve for time should be straightened up straightforward enough. So the first thing asks is just create a free body diagram. So this is our toolbox or on the bed of the truck for saying we're traveling to the right hand sign to be positive extraction and up is positive line, so that for extending to the right in this scenario, then our Well, then, I guess for selling to the right, then the box will wanna move backwards. So our force of static friction will be acting in this direction. So a here represents the motion of the truck. That means the then say that the toolbox want trouble backwards. So of course, this act friction is acting to the right. And of course, we have a number for us acting upward and balance by the weight of truck. It's that the toolbox itself. So for this scenario, look at the balance of forces in our X direction. We know that the track itself is moving with a and every ad in our coefficient of her sari or static friction for us is equal zero. The acceleration is simply our native value for the, uh, fiction for us, which is in us times n you really get forces in our wind direction. But you're also is, you know, then we have the normal force must people the weight of of the toolbox and so we can replace our value. Sorry for acceleration. So this is over. Massa's well forgot to name us times the weight which is sometimes g divided back by the mass of the toolbox. We can cancel out our values for a mass. We can find that our term for acceleration is native. All this pleasant 0.6 fire times G, 9.8 meters per second squared using us, remember, because through the two boxes, initially not moving. So I had to overcome the force of static friction over For this Dacre, we find that acceleration would be negative. Who hadn't checked? It actually looks irritable. Data fix everything is quite 65 times 9.8, 6.37 meters per second squared needed because if the trunk is traveling to the rate and of course the toolbox will go in the opposite direction. That's why that makes sense. If then we want to find the velocity, which is, uh, those sort of time looking for the short time for the scenario to occur. And if we know where Final velocity, which is the last day of the truck we started from initial velocity plus 80 for starting again from rest. The loss evildoing will be negative. Actually, the velocity of the two blocks will be the opposite direction. So you find the time is simply the negative velocity over the acceleration. Just a native 30 meters per second. I'm a negative acceleration of 6.37 meters per second squared, which results any time of 4.71 seconds. How is this question hinted to US? Means uses Newton's second law to solve for the acceleration off the system the truck In this case, I found acceleration of the of the toolbox in the negative X direction which results in the truck and is the trunk of a velocity of 30 meters per second and the two lakhs having a negative velocity of 30 meters per second, such that we can solve for the final time, using both these velocities and accelerations. So there you go.

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