Question

Bridging Problem: A Spring and Friction on an Incline 27 of 27 Review Constants Part 1 A package of mass 600 klased on a 531'incine, a distance D-500m from a long spring with force constant 1.10 x 10 N/ms that is tached the bottom of the nine Figure 1) The coeficients of thiction between the package and incin 0.400 and 0.200. The mass of the spring is negotij What is the maximum compression of the song The package rebounds up the incine. When it stops again how close does it get to song position () What is the change in the internal energy of the package and incine from the point at which the package is released un rebounds to madun her Calculate the instant not force cented on the package at its lowest point is the package remain there? Express your answer with the appropriate units. Recall that down the incline is the direction and up the incline is the direction View Available NA 2 Value Units Submit Figure Part 1 Using the general energy equation for the motion of the package between the lowest and the final points. Ind how close the package is to sorginal position Express your answer with the appropriate units НА ? Value Units Submit

Answer 1

When the package is at the highest point it has gravitational potential energy. And when it is released its energy going to be used in two ways.

1. Some energy get dissipated due to kinetic frictional force acting on it.
2. Some part of energy get stored in spring as spring potential energy.

So let's first make the free body diagram

When package has not touched the spring

Diagram 1:

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Here R=normal reaction and f_{​​​​​​k} =kinetic frictional force.

Now suppose the maximum compression in the spring is x when package reached its lowest point. Then at this position the package comes to instantly rest and therefore when spring tends to bounce back the static frictional force starts acting at the same instant.

Diagram-2(a):

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Here f_{​​​​​​s} =static frictional force and f_{​​​​​​sp} = spring force.

Diagram-2b:

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So from diagram-2(b) and diagram-1 the maximum gravitational potential energy stored in the package with respect to its lowest point is

E = mg D +.) sin ........... (1)

And energy dissipated due to kinetic frictional force is equal to work done by it which is equal to

W = (D+rfk...... (2)

But

fk = MKR...........(3

To find the value of R apply Newton's law of motion along y-axis in diagram-1 or 2(a) we get

R mg cos

Put this value in equation (3) and then value of (3) in (2) we get

1 ... soo burl(x + a) = M

And at its maximum compression the energy stored in the spring is

Ek kr?....... .(5)

Therefore by conservation of energy sum of equations (4) and (5) is equal to equation (1). Therefore

mg(D + 2) sin 8 = 3kir+ (D + x)Hảmg cos e 1 +0=-kx2 + (D+2) mg cos 0 – mg(D+ x) sin 0 ka? → kr? + 2.xmg uk cos - sin ) + 2Dmg(ucos - sin ) = 0.............. 2 (6)

Put in equation (6) given values

k = 1.1 x 10-N/m, m n = 6.0kg, Mk = 0.20, 0 = 53.1°, and g = 9.8m/s2

And D=5.0m

We get

→ 110.x2 – 79.92.1 – 399.60 = 0........... ....(7)

We can see equation (7) is quadratic equation so solved it with quadratic formula we get

x = 2.30m or I= -1.57m

But negative value ruled out therefore maximum compression in the spring is 2.39m

Now at the instant when package is at its lowest point the acting forces on it is shown in diagram-2(a)

At this instant

Apply Newton's law of motion we get

Along y-axis

R= mg cos ..... (8)

Along x-axis

F=fs + mg sin 8 – tsp F = pyR+ mg sin - kr..........(9)

Put value of R from (8) we get

$F=\mu_smg\cos\theta+mg\sin\theta-kx\\ \\\Rightarrow\ F=mg(\mu_s\cos\theta+\sin\theta)-kx..........(10)$

Put all the values we get

$F=6.0X9.8(0.4X\cos 53.1\degree+\sin 53.1\degree)-110X2.3\\ \\\Rightarrow\ F=-191.85N$

This is the force acting on the package at its lowest point. Here its negative sign shows that net force acting on the package is upward along x-axis therefore package will start moving upward along x-axis.

Now at the lowest point the whole energy of the system is stored in the spring in the form of spring potential energy given by equation (5).

When the package starts moving upward along x-axis some part of this stored energy get dissipated due to kinetic frictional force acting on the package and some part of it get stored in the package as gravitational potential energy.

Now let's assume the maximum height reached out by the package along x-axis is L. See the figure

20 D+4 sino freel -L Sino

Then energy dissipated due to the kinetic frictional force is

$W=f_kL=\mu_kRL=\mu_kmg\cos\theta L.........(11)$

And gravitational potential energy stored in the package at its highest reached point is

$E_p=mgL\sin\theta.........(12)$

Therefore by conservation of energy

$E_k=W+E_p$

Using equation (5), (11) and (12) we get

$\frac{1}{2}kx^2=\mu_kmg\cos\theta L+mgL\sin\theta\\ \\\Rightarrow\ L=\frac{kx^2}{2mg(\mu_k\cos\theta +\sin\theta)}$

Put all the values we get

$L=\frac{110X2.3^2}{2X6.0X9.8(0.2\cos 53.1\degree +\sin 53.1\degree)}\\ \\\Rightarrow\ L=5.4m$This is the maximum height reached out by the package along x-axis.

therefore its closeness with its initial highest position is

$D+x-L=5.0+2.3-5.4=1.9m$

this is the closeness length.

Change in internal energy of the package is

$\Delta E=mg(D+x)\sin\theta-mgL\sin\theta$

Or

$\Delta E=6.0\times9.8(5.0+2.3)\sin53.1\degree-0.6\times9.8\times5.4\sin53.1\degree\\ \\\Rightarrow \ \Delta E=83.34J$thos is the change in internal energy of the package .