(III) A spring ( k = 75 N/m) has an equilibrium length of 1.0...

Question

(III) A spring ( k = 75 N/m) has an equilibrium length of 1.00 m. The spring is compressed to a length of 0.50 m and a mass of 2.0 kg is placed at its free end on a frictionless slope which makes an angle of 41° with respect to the horizontal (Fig. 8–41). The spring is then released.

(c) Now the incline has a coefficient of kinetic friction μₖ . If the block, attached to the spring, is observed to stop just as it reaches the spring’s equilibrium position, what is the coefficient of friction μₖ ?

Accepted Answer

Hello, let's go through this practice problem. A spherical ball of mass 2.5 kg is attached to one end of a spring. The other end of the spring is fixed at the foot of an inclined plane. As shown in the figure, the incline makes an angle of 32 degrees with the horizontal, the spring with an equilibrium length of 1.6 m is compressed to 0.8 m and then released afterward. The ball traveled up the incline and stopped just as it reached the spring's equilibrium position. If the coefficient of kinetic friction between the ball and the inclined surface is new, calculate its value assume the spring constant is 62 new ones per meter option. A 0.33 B 0.57 C 0.87 or D 1.8. So because we have a system here that involves changes in speed changes in gravitational potential energy and changes in elastic potential energy. We can solve this problem using the law of conservation of energy which states that the initial and final amounts of mechanical energy of the system should be equal to one another. So another way of writing this is to say the change in each component of energy will add up to zero. There are a few different sources of energy. So one way to write this is to say that the change in kinetic energy plus the change in potential energy plus the change in work due to the fact that friction is doing some work on the ball. So plus the work from the friction, remember that the work is equal to force multiplied by the distance over which that force is acting. So I'll say F sub f the force from the friction multiplied by, we'll call it s the distance over which the force acts. This sum of changes in energy should add up to zero. So let's expand these equations out using formulas. We know recall that kinetic energy is equal to one half multiplied by the mass multiplied by the square of the speed. So if representing that as a change, that is one half multiplied by the difference between the squares of the speeds. So we'll call the final squared minus the initial squared. That's our kinetic energy term. Now for the potential energy, remember there are two different components of potential energy. There's the elastic potential energy from the spring and the gravitational potential energy from the gravity from vertical height. So we'll use the same thing. So for elastic potential energy, the formula is one half K multiplied by the square of the compression distance of the spring So one half K multiplied by, we'll call it delta X final squared minus delta X initial squared plus the change of potential energy. Again, we'll use a similar process. Potential energy is equal to mass multiplied by the gravitational acceleration. G multiplied by the change in height. So that's the mass multiplied by G multiplied by Y final minus Y initial. And then finally, plus the work done by the friction. So recall that frictional force is equal to the coefficient of friction mu multiplied by the normal force, which is equal to the weight of the, of the, of the object. In this case, MG multiplied by the distance which since it said an angle is going to be s multiplied by the cosine of the angle phi and all this is equal to zero. But our goal in the problem is to solve this equation for me, for the, for the coefficient of friction. So we'll do that using some algebra. But first, there are a few ways we can simplify this equation first off since the initial and final states given by the problem are both sp we're specifically told that both of those states are states where the ball is not moving where VF and V I are both equal to zero, which means we can completely cross out the kinetic energy term. The final compression of the spring delta X sub F squared can also go to zero because the, we're told that the ball stops at the equilibrium position. And the X is defined as the distance from the equilibrium position. So that variable is just going to be zero. Also for the potential energy term. Remember that in problems with potential energy, it's most helpful for us to treat the initial Y term or at least the lower Y term as though it's at zero to make our calculations more simple or Y sub F is just going to be equal two with that s distance multiplied by the sine of the angle of the wedge. So with all that in mind doing some pretty basic algebra on this tells us that mu is equal to negative one half K delta X sub I squared plus MGS sign five all divided by negative MGS cosine phi we can simplify this equation a little bit just to make it easier for us to type in our numbers by splitting up the numerator because we've got a plus sign here. So we can split those terms up into two separate terms where we'll notice that the sign in the numerator can be combined with the cosine in the denominator to create a second term where we just have a tangent of phi. So the equation can more simply be written as K delta X of I squared divided by two and GS cos minus the tangent of fire. And now we can just plug our values into a calculator. So the spring constant is 62 newtons per meter multiplied by the square of delta X. So 0.8 m divided by two multiplied by the mass of the ball 2.5 kg multiplied by the gravitational acceleration. G 9.81 m per second squared multiplied by S 0.8 m un multiplied by the cosine of the angle 32 degrees. And if you put that into a calculator and then minus the tangent of 32 degrees. And if we put that into a calculator, and we find a value for me of about 0.5675 which can be rounded to about 0.57. And so that is our answer which corresponds to option B. So option B is our solution and that's it for this problem. I hope this video helped you out and please consider checking out our other tutoring videos which will give you more experience with these types of problems. That's all for now. And I hope you have a lovely day. Bye bye.

Key Concepts

Hooke's Law

Friction and Coefficient of Friction

Inclined Plane Dynamics

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