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
