A 1.0 kg block is attached to a spring with spring constant 16 N/...

Question

A 1.0 kg block is attached to a spring with spring constant 16 N/m. While the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 40 cm/s. What are

b. The block's speed at the point where 𝓍 = ½A?

Accepted Answer

Hey, everyone in this problem, we're asked to consider a spring of spring constant 12 newtons per meter connected to a 2.5 kg block. The block is initially at rest but then a force of 15 newtons is applied to it imparting a speed of 25 centimeters per second to the block. We are asked to determine the blocks speed when it reaches a displacement of X equals 0.3 m. We're given four answer choices. All four of them have the unit of meters per second. Option A is 0.51. Option B is 0.38. Option C is 0.24 and option D is 0.58. Now let's draw a little diagram of what we have here. So we have a spring. In this case, we're drawing a horizontal spring and we have a 2.5 kg block connected to it. And our spring constant K is 12 noons per meter sometime later. OK. Our spring is gonna be compressed of it and our block is gonna be a distance of 0.3 m from its original position. OK? We have this displacement of 0.3 m. Now we're told that this force is applied which imparts a particular speed. OK? And that particular speed is gonna be V knot. OK. Our initial speed and that is corresponding with that first diagram. As soon as that force is applied, we get this speed V knot of 25 centimeters per second. Let's convert to our standard unit. So we're gonna multiply it by one m divided by 100 centimeters. OK? We know that there are 100 centimeters in every meter. So the numerator and denominator are equivalent. It's like we're multiplying by one, the unit of centimeter is going to divide out. And what we're doing is essentially dividing by 100 to get 0.25 m per second. And what we want to find is the speed V F at this second point in time. OK? We only have this displacement of 0.3 m. Now, let's think about our conservation of mechanical energy. OK? We aren't told anything about friction here. So we're assuming that this is frictionless. We don't have any external forces acting. So let's write down our conservation of mechanical energy. We have K knot, the initial kinetic energy plus the UK, not the initial potential energy is going to be equal to K F, the final kinetic energy plus U F the final potential energy. Now, in this case, we have a horizontal spring. So we have no um gravity acting, we have no potential energy due to gravity. And so the potential energy is gonna be made up solely of the potential energy due to its spring. OK. Recall that kinetic energy is given by one half M V squared. Potential energy from the spring is given by one half K X squared. And so our equation becomes one half M V not squared plus one half K X not squared is equal to one half M V F squared plus one half K X F squared. Now, initially, this block is at rest. OK. That means that the spring is gonna be in its equilibrium position when that speed is imparted. So X knot is actually gonna be zero. The spring potential energy initially is therefore going to be zero. So on the left hand side, all we have is one half multiplied by the mass of 2. kg multiplied by that initial speed, 0.25 m per second squared. And this is gonna be equal to one half multiplied by the mass 2.5 kg multiplied by the final speed V F squared plus one half multiplied by the spring constant, 12 newtons per meter multiplied by the displacement 0.303 m squared. Now we have an equation with just one unknown. That's V F. That's what we're trying to solve for. So all that's left to do is simplify and solve for V F. Simplifying on the left hand side, we have 0. kg meters squared per second squared. On the right hand side, we have 1.25 kg multiplied by V F squared, 0. Newton meters. Hm. Now again, we wanna isolate for V F. So we're gonna move the 0.54 Newton meters to the left hand side. Now recall that a Newton is equivalent to a kilogram meter per second squared. So these units of kilogram meter squared per second squared and new meters are equivalent. OK? So we can add those together. No problem. We have 1.25 kg multiplied by V F squared, going to be equal to 0.72725. And we're gonna write this as kilogram meter squared per second squared. OK? Dividing by 1.25 kg V F squared is gonna be equal to 0. m squared per second squared. And finally taking the square root, we get that the final speed is gonna be 0.2412 m per second. Now, when we take the square route, it's important to remember that you get both the positive and the negative route. In this case, we're looking for just the speed. So we want just the magnitude we don't care about the direction. So we take the positive route um just for that purpose. And so that is gonna be the final speed or the speed of the block when it reaches that displacement of 0.3 m. If we go up to our answer choices and compare what we found, we can see that these are rounded to two significant digits and the answer we found corresponds with answer choice C 0.24 m per second. Thanks everyone for watching. I hope this video helped see you in the next one.

A 1.0 kg block is attached to a spring with spring constant 16 N/m. While the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 40 cm/s. What are b. The block's speed at the point where 𝓍 = ½A?

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