A horizontal spring with spring constant 250 N/m is compressed by...

Question

A horizontal spring with spring constant 250 N/m is compressed by 12 cm and then used to launch a 250 g box across the floor. The coefficient of kinetic friction between the box and the floor is 0.23. What is the box's launch speed?

Accepted Answer

Hey, everyone in this problem, a ball of 0.5 kg initially at rest is pushed horizontally by a spring with a spring constant of 100 newtons Per meter. The sprint is compressed by 20 cm before pushing the ball. If the coefficient of kinetic friction is 0.27, what would be the final speed of the ball as it leaves the contact of the spring? We're given four answer choices. We have a 6.4, m/s. B 2.3, m/s. C 6.94 m/s and d 2.63 m/s. All right, we're looking for the spinal speed. Hey, we're given information about the kinetic friction about the spring constant, the compression. Let's recall that the total work is equal to the change in kinetic energy. Delta K hm. So if we can find the work done here, yeah, we can relate to, to, to the kinetic energy which includes a value for the final speed. Ok. So let's try to start there. Now, in this case, what is the work total? Ok. Well, we have two things to consider we have the work done by the spring. We're also told that there is kinetic friction. And so we also have to worry about the work done by friction. So we get the work of the spring plus the work of the friction is equal to the change in kinetic energy delta K. All right, let's start with the work done by the spring. The work done by this spring, it's going to be equal to one or recall that we can make the work as the integral from the initial position to the final position of the force in the X direction D X. Now, when we're talking about a spring, we'll call it, the sprint force is equal to K multiplied by X. So the integral, now we're going from initial resting position Where X is equal to 0cm up to a position. X is equal to 20 centimeters because the spring is compressed centimeters. So we have the integral from X equals zero centimeters to X equals 20 centimeters of K multiplied by X D X. And when we take the integral of K multiplied by X, with respect to X, we raise the power by one divide by that new power. So we get one half multiplied by K multiplied by X squared. And this is gonna be evaluated at X equals zero, OK. Meters. Now, we have X equals 20 centimeters. We want to convert that to our standard unit of meters. We're gonna do that up here. So we have X is equal to 20 centimeters. We're gonna multiply by one m For cm. Whoops, not per 20 cm per 100 cm. OK? We have 100 cm in every meter. The unit of centimeters will divide. OK? And we're essentially taking our centimeters 20 dividing by 100 to get our meters. 0.2 m. OK? So now we're evaluating from X equals zero m to X equals 0.2 m. And you'll recognize this expression one half multiplied by K multiplied by X squared. That's the energy of the spring. OK. The spring potential energy. OK. So what we're doing when we're finding the work is we're essentially finding the change in spring energy from these two positions. Now we end up with one half, the spring constant is 100 newtons per meter Multiplied by 0.2 m squared. OK? And then we're gonna subtract the evaluation at the lower bone. In this case, it just gives us a value of zero since X is zero. So we have one half multiplied by 100 multiplied by .2 m squared. And we end up with a value of two for our units, we have a new per meter multiplied by meter squared. So we're gonna get a newton multiplied by a meter and that is equivalent to a jewel. So we have two jewels. So that is our spring. The work done by the spring. That's the first value we were looking for. Now, let's find the work done by friction. OK. So we can use those in our equation with our kinetic energy to get the final speed. So the work done by friction I recall is going to be equal to the force due to friction multiplied by the distance it acts over s cosine of theta where theta is the angle between the force and the emotion and our force of friction. And this is gonna be equal to UK coefficient of kinetic friction multiplied by the mass M multiplied by the gravitational acceleration. G multiplied by D multiplied by coast data. I have a whole bunch of things to multiply together here. Now substituting our values, we have the coefficient kinetic friction of 0. Multiplied by a mass of 0.5 kg multiplied by gravitational acceleration 9.8 m per second squared Multiplied by the distance. OK. Which is how much that spring is compressed by which is .2 m Multiplied by cosine of the angle. Now, we know that friction is going to act opposite to the motion. OK. So the angle is gonna be 100 degrees indicating that they're going in opposite directions, cosine of 100 gives us a negative value as we expect the work done by friction to be negative. And we get a value Of -0.2, 6, 46 Jews. All right. So Now, if we look back at this work done by the spring plus the work done by friction is equal to the change in kinetic energy delta K. We have the following. Let me rewrite that equation, but we can see it still on our page. We can still refer to it. All right. Now, the work done by the spring that we found was two jewels. The work done by friction was negative 0. 46 jewels. So we have two jewels minus 0. jewels. And this is gonna equal the change in kinetic energy which is equivalent to one half M V F squared minus one half multiplied by M multiplied by V not squared. I recall that kinetic energy is given by one half M V squared. And so this is a change the final minus the initial On the left hand side, we can simplify to 1.7, s And on the right hand side, the initial speed is zero. So the second term goes to zero, we only have to worry about that first term where we have our final speed that we're looking for. We have one half multiplied by a mass of 0.5 kg multiplied by V F squared. And we're looking for V F, we can rearrange, we're gonna divide both sides by one half multiplied by 0.5 kg. We have the V F squared is equal to 1.7354 jules divided by 0.25 kg. If we work this out on our calculator, we get that V F squared is equal to 6. meter squared per second squared. Ok. We're called that a Jeel is a kilogram meter squared per second squared. So when we divide that by kilogram, we're left with meter squared per second squared. When we take the square route, we're gonna get both the positive and negative route. In this case, we're looking for the speed. So we only care about the magnitude. So we're just gonna write down the positive route and that is gonna be 2.6347m/s, right? And that is gonna be the final speed of the ball as it leaves contact with the spring. If we go back up to our answer choices, remember that they were rounded to three significant digits. So if we round the answer, we found to three significant digits, we see that we have 2.63 m per second, which corresponds with answer choice. D Thanks everyone for watching. I hope this video helped see you in the next one.

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A horizontal spring with spring constant 250 N/m is compressed by 12 cm and then used to launch a 250 g box across the floor. The coefficient of kinetic friction between the box and the floor is 0.23. What is the box's launch speed?

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