A 2 kg package is released on a 53.1° incline, 4 m from a long spring with force constant k = 140 N/m that is attached at the bottom of the incline (Fig. 7-32). The coefficients of friction between the package and the incline are µs = 0.4 and µk= 0.2. The mass of the spring is negligible.A. What is the speed of the package just before it reaches the spring?

B. What is the maximum compression of the spring?

C. The package rebounds back up the incline. How close does it get to its initial position?

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Answer:

A) The speed of the package just before it reaches the spring = 7.31 m/s

B) The maximum compression of the spring is 0.9736m

C) It is close to it’s initial position by 0.57m

Explanation:

A) Let’s talk about the motion;

As the block moves down the inclines plane, friction is doing (negative) work on the block while gravity is doing

(positive) work on the block.

Thus, the maximum force due to

static friction must be less than the force of gravity down the inclined plane in order for the block to slide down.

Since the

block is sliding down the inclined plane, we’ll have to use kinetic friction when calculating the amount of work

(net) on the block.

Thus;

∆Kt + ∆Ut = ∆Et

∆Et = ∫|Ff| |ds| = – Ff L

Where Ff is the frictional force.

So ∆Kt + ∆Ut = – Ff L

And so;

(1/2)m((vf² – vo²) + mg(yf – yo) = – Ff L

Resolving this for v, we have;

V = √(2gL(sinθ – μkcosθ)

V = √(2 x 9.81 x 4) (sin53.1 – 0.2 cos53.1)

V = √(78.48) (0.68))

V = √(53.3664)

V= 7.31 m/s

B) For us to find the maximum compression of the spring, let’s use the change in kinetic energy, change in

potential energy and the work done by friction.

If we start from the top of the incline plane, the initial and final kinetic energy of the block is zero:

Thus,

∆Kt + ∆Ut = ∆Et

And,

∆E = −Ff ∆s

Thus;

mg(yo – yf) + (k/2)(∆(sf)² – ∆(so)² = −Ff ∆s

Now let’s solve it by putting these values;

yf − y0 = −(L + ∆d) sin θ; ∆s = L + ∆d; ∆sf = ∆d; and ∆s0 = 0 where ∆d is the maximum compression in the spring.

So, we have;

((1/2

)(K)(∆d

)²) − ∆d (mg sin θ − (µk)mg cos θ) + ((µk)mgLcos θ − mgLsin θ) = 0

Let’s rearrange this for easy solution.

((1/2)(K)(∆d)²) − ∆d (mg sin θ − (µk)mg cos θ) – L(mgsin θ – (µk)mgcos θ) = 0

Divide each term by (mgsin θ – (µk)mgcos θ) to get;

[((K/2)(∆d)²)}/{(mgsin θ – (µk)mgcos θ)}] – ∆d – L = 0

Putting k = 140,m = 2kg, µk = 0.2 and θ = 53.1° and L=4m, we obtain;

5.247(∆d)² – ∆d – 4 = 0

Solving as a quadratic equation;

∆d = 0.9736m

C) let’s find out how high the block rebounds up the

inclined plane with the fact that final and initial kinetic energy is zero;

mg(yf − yo) + 1

/2

k (∆s

f² − ∆s

o²) = −Ff ∆s

Now let’s solve it by putting these values; yf − y0 = (L′ + ∆d)sin θ; ∆s = L′ + ∆d; ∆sf = 0; and ∆s0 = ∆d.

L’ is the distance moved up the inclined plane

So we have;

(1/2)k∆d² + mg(∆d + L′)sin θ =

-(µk)mg cos θ (∆d + L′)

Making L’ the subject of the formula, we have;

L’ = [(1/2)k∆d²] /(mg sin θ + (µk)mg cos θ)] – ∆d

L’ = [(140/2)(0.9736²)] /(2 x 9.81 sin51.3) + (0.2 x 2 x 9.81cos 53.1)] – 0.9736

L’ = (66.353)/[(15.696) + (2.3544)]

L’ = (66.353)/18.05 = 3.43m

This is the distance moved up the inclined plane. So it’s distance feom it’s initial position is 4m – 3.43m = 0.57m