(III) A spring (k=75 N / m) has an equilibrium length of 1.00 m . The spring is compressed to a

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

(III) A spring $(k=75 \mathrm{N} / \mathrm{m})$ has an equilibrium length of 1.00 $\mathrm{m} .$ The spring is compressed to a length of 0.50 $\mathrm{m}$ and a mass of 2.0 $\mathrm{kg}$ is placed at its free end on a frictionless slope which makes an angle of $41^{\circ}$ with respect to the horizontal (Fig. $38 ) .$ The spring is then released. (a) If the mass is not attached to the spring, how far up the slope will the mass move before coming to rest? (b) If the mass is attached to the spring, how far up the slope will the mass move before coming to rest? (c) Now the incline has a coefficient of kinetic friction $\mu_{k}$ . 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 $\mu_{k} ?$

(III) A spring $(k=75 \mathrm{N} / \mathrm{m})$ has an equilibrium length of
1.00 $\mathrm{m} .$ The spring is compressed to a length of 0.50 $\mathrm{m}$ and a
mass of 2.0 $\mathrm{kg}$ is placed at its free end on a frictionless slope
which makes an angle of $41^{\circ}$ with respect to the horizontal (Fig. $38 ) .$ The spring is then released. (a) If the mass is not attached to the spring, how far up the slope will the mass move before coming to rest? (b) If the mass is attached to the spring, how far up the slope will the mass move before coming to rest? (c) Now the incline has a coefficient of kinetic friction $\mu_{k}$ . 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 $\mu_{k} ?$

Key Concepts

Inclined Plane Mechanics

Inclined plane mechanics involves analyzing the motion of objects on a slope by resolving forces into components parallel and perpendicular to the surface. This approach is critical in understanding how gravitational force contributes to the motion along the incline, as well as determining normal forces relevant to frictional calculations.

Conservation of Mechanical Energy

This principle states that in a closed system with only conservative forces acting, the total mechanical energy (the sum of kinetic and potential energies) remains constant. In the context of the problem, it is used to equate the energy stored in the spring with the gravitational potential energy gained as the mass moves along the incline.

Friction and Energy Dissipation

Friction, particularly kinetic friction, is a non-conservative force that opposes motion and converts mechanical energy into thermal energy. Its effect is quantified by the coefficient of kinetic friction in conjunction with the normal force. Recognizing the work done by friction is essential when energy is not fully conserved due to such dissipative forces.

Elastic Potential Energy

This concept involves the energy stored in a spring when it is either compressed or extended. It is quantified by the expression ½ k x², where k is the spring constant and x is the displacement from the equilibrium position. This energy can be converted into other forms, such as gravitational potential energy, when the spring is released.

Gravitational Potential Energy

Gravitational potential energy represents the energy an object possesses due to its position relative to a gravitational field. It is given by the product mgh, where m is mass, g is the acceleration due to gravity, and h is the vertical displacement. This concept is fundamental when considering the work done against gravity on an inclined plane.

Transcript

00:01 All right, we're using conservation of energy throughout this problem.

00:05 And essentially you have spring force, or elastic potential energy of the spring, gravitation of potential energy, and no kinetic energy.

00:16 So in the first part, part a, you're looking at when the block is not attached to the spring, looking at initial potential energy or spring energy, elastic potential energy, to be equal, and you said that equal to gravitational potential energy, the top or at whatever height reaches.

00:45 And so initial elastic potential energy is one -half kx -1 squared, x -1 being the initial displacement of 0 .5 meters.

00:55 Gravitational potential energy that this is converted to is mg times d sine theta.

01:03 And the reason it's the reason where, well, this is y and y has to be, well, sign theta is y over d.

01:13 Right.

01:17 So this d is what we're interested in.

01:20 Therefore, d sine theta will give us y, which is why this turned out the way it therefore, d is what we're after is, is 1⁄2x1 squared over 2mg times sine theta.

01:43 Okay? and k is 75 newton's per meter, x1 is 0 .5 meter quantity squared over 2 times 2 kilograms times 9 .8.

01:56 Meters per second squared times sine theta which is 41 giving us a d of 0 .73 meters.

02:08 In part b we have initial elastic potential energy is equal to gravitational potential energy at the top let's call ug2 plus the elastic potential energy at the top so u.

02:28 U2.

02:28 Okay, in other words, 1 half k x1 squared is equal to again mgd sine theta plus one half k times extension squared.

02:41 So this would be d minus in an x1 quantity square.

02:45 So let's plug in some values, make things easier before we solve the quadratic.

02:50 So this is one half times 75 times 0 .5 squared is equal to 2 times 9 .8.

03:01 Times .73 times .41 plus 1⁄2x75 times d minus .5 square, whole square.

03:19 All right, so looking at the quadratic on the right hand side, you have 37 .5d squared.

03:31 Minus 37 .5d, that's what the second term from this will give us, plus 1 .75 times 0 .5 squared, which is the same of this term on the left, so those terms cancel...

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