**Solved Physics Homework Problems on Mechanical
Energy Conservation.**

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**Example One, Homework Help, Problem Solution:
Conservation of Energy Applied to a Case of a Spring Compression.**

**A ball of mass m is placed at a distance
h above the end of a vertical spring. The ball is then released and compresses
the spring. The elastic constant of the spring is k. What is the maximum spring
deformation? Assume no friction.**

**The image above shows the initial and final
situations for this problem.**

**It is convenient to choose the lowest point reached by the ball as the reference level (y = 0) . We could choose a point such as point A, but it would have the disadvantage of having to deal with positive and negative potential energies****.**

**Let's notice that in the final situation
the ball is momentarily stop and its kinetic energy becomes zero at the stop
point.**

**The following table help us to explain the
solutions steps:**

** Initial
situation Final situation**

Kinetic Energy Ec_{i}
= 0 Ec_{f}
= 0

Gravitacional Potential Energy Epg_{i}
= mg(h + y) Epg_{f}
= 0

Elastic Potential Energy Epe_{i}
= 0 Epe_{f}
= ½ ky^{2}.

**Then, as the initial energy must equal the
final energy,**

**(Ec + Epg + Epe) _{i} = (Ec + Epg
+ Epe)_{f}, we obtain:**

**mg(h
+ y) = ½ ky ^{2}.**

**The maximum deformation or compression of
the spring is obtained solving for y in this quadratic equation. The parameters
m, h and k are the known data.**

**For example, suppose a ball of mass m =
0,4 kg is dropped from a height h = 1.2 m and an the elastic constant k =
100 N/m. Replacing this data and solving the equation we get y = 0.35 m or
35 cm.**

**Example Two, Homework Help, Problem Solution:
Conservation Of Energy Applied to a Vehicle Braking Within a Distance.**

**An automobile travels with a speed of 72
km/h. Assume that when the brakes are fully pressed, it can be stopped on
a 25 meters distance. If the vehicle had double velocity, 144 km/h, what will
its braking distance be?. Let's also assume that the friction force between
tires and ground is nearly constant. Also calculate the coefficient of friction.**

**Solution:**

** We can use "work done on an object is equal
to its kinetic energy change" , W = Change in Ek = Ek
= Ekf - Eki. W is the net work done on the object. As the force is supposed
to be constant, the work done by the friction force F is Fd, where d is the
braking distance. F and d are in opposite directions, so W is negative. The
final velocity is zero. Then **

**
W = - Fd = Ek = 0 - ½
mv ^{2}.**

**d
= ½ mv ^{2}/F.**

**m and F are constants, so it is clear that
the braking distance increases proportionally with the square of velocity.
**

**The braking distance for the 144 km/h velocity
is 100 meters, four times the braking distance needed for 72 km/h.**

**To calculate the coefficient of friction,
we use Fd = ½ mv ^{2}.**

** F= µmg, µ is the friction coefficient
and m is the vehicle mass.**

**µmgd
= ½ mv ^{2}.**

**The velocity is 72 km/h = 72/3.6 m/s = 20
m/s and d is 25 m. **

**Then µ
= ½ v ^{2}/gd = ½ (20 m/s)^{2}/(9.8 m/s^{2} 25 m) =
0.816 .**

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**Sites With Solved Exercises on Mechanical
Energy **

** · Physics
Homework - Examples of Mechanical Energy Problems
**

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