# Mechanical energy – problems and solutions

**The work-mechanical energy principle**

1. The coefficient of the kinetic friction between block and floor (μ_{k}) is 0.5. What is the displacement of an object (s)? Acceleration due to gravity is 10 m/s^{2}.

__Known :__

The coefficient of the kinetic friction (μ_{k}) = 0.5

Mass of block (m) = 4 kg

Acceleration due to gravity (g) is 10 m/s^{2}

Weight of block (w) = m g = (4)(10) = 40 Newton

If block on the horizontal plane then the normal force (N) =weight (w) = 40 Newton.

If block on the horizontal plane then the normal force (N) = weight (w) = 40 Newton.

Initial velocity (v_{1}) = 5 m/s

Final velocity (v_{2}) = 0 m/s

__Wanted :__ displacement of object (d) ?

__Solution :__

The work-mechanical energy principle states that work (W) done by the nonconservative force is the same as the change of the mechanical energy of an object. The change of mechanical energy = the final mechanical energy – the initial mechanical energy.

The kinetic friction force is one of the nonconservative forces and the only one nonconservative force that acts on the block.

W = ΔME

f_{k }d = ME_{2} – ME_{1}

Work done by the kinetic friction force :

W = f_{k} d = (μ_{k})(N)(d) = (0.5)(40)(d) = 20 d

The change of the mechanical energy :

ΔME = ME_{2} – ME_{1} = (KE + PE)_{2} – (KE + PE)_{1}

Object moves along the horizontal plane and no change of height (Δh = 0) so there is no change of the gravitational potential energy

_{2}– PE

_{1}= 0). Thus the change of the mechanical energy just involves the change of the kinetic energy.

ΔME = KE_{2} – KE_{1} = ½ m v_{2}^{2} – ½ m v_{1}^{2 }= ½ m (v_{2}^{2 }– v_{1}^{2})

**Δ****M****E**** = ½ (4)(0**^{2}** – 5**^{2}**) = (2)(25) = 50**

__Displacement of block :__

W = ΔME

20 s = 50

s = 50 / 20

s = 2.5 meters

**The principle of conservation of mechanical energy**

2. Object A and object B have the same mass. Object A free fall from a height of h meters and object B free fall from a height of 2h meters. If object A hits the ground at v m/s, then what is the kinetic energy of the object B when it hits the ground.

__Solution :__

The final velocity of object B when free fall from a height of 2h :

v^{2} = 2 g (2h) = 4 g h

The kinetic energy of object B :

KE_{B} = ½ m v^{2} = ½ m (4 g h) = 2 m g h —– equation 1

The initial mechanical energy of object B = the gravitational potential energy = m g h.

The final mechanical energy of object B = the kinetic energy = ½ m v^{2}.

The principle of conservation of mechanical energy :

m g h = ½ m v^{2}

Because __m g h = ½ m v__^{2} then we can change __m g h__ in equation 1 with ½ m v^{2}.

The kinetic energy of object B = 2 m g h = 2(½ m v^{2}) = m v^{2}

3. An object free fall from a height of 20 meters. Acceleration due to gravity is 10 m/s. What is the velocity of object 15 meters above the ground?

__Solution :__

The final mechanical energy = the initial mechanical energy

The kinetic energy at point 2 = the change of the gravitational potential energy as far as 5 meters.

4. A block is released from the top of the smooth inclined plane. What is the velocity of the block when hits the ground?

Solution :

The initial mechanical energy = the gravitational potential energy = m g h = m (10)(5) = 50 meters

The final mechanical energy = the kinetic energy = 1/2 m v^{2 }

The principle of conservation of mechanical energy, states that the initial mechanical energy = the final mechanical energy.

ME_{o} = ME_{t}

50 m = 1/2 m v^{2 }

50 = 1/2 v^{2 }

2 (50) = v^{2 }

100 = v^{2 }

v = 10 m/s

5.

A block with mass of m-kg released from a height of h meters above the ground, as shown in figure below. Determine the ratio of the potential energy to the kinetic energy (KE) at point M.

__Solution :__

The gravitational potential energy at point M :

PE_{M} = m g (0.3 h)

The kinetic energy at point m = the change of the gravitational potential energy as far as h-0.3h = 0.7 h

KE_{M} = PE = m g (0.7 h)

The ratio of the gravitational potential energy to the kinetic energy at point M :

PE_{M} : KE_{M}

m g (0.3 h) : m g (0.7 h)

0.3 : 0.7

3 : 7

6. If PE_{Q} and KE_{Q} have the potential energy and the kinetic energy at point Q (g = 10 m/s^{2}), then PE_{Q} : KE_{Q} =…

__Solution :__

The gravitational potential energy at point Q :

PE_{Q} = m g h = (m)(10)(1.8) = 18 meters

The kinetic energy at point Q = the change of the gravitational potential energy as far as 5-1.8 = 3.2 meters

KE_{Q} = PE = m g h = m (10)(3.2) = 32 m

The ratio of the gravitational potential energy to the kinetic energy at point Q :

PE_{Q} : KE_{Q }

18 m : 32 m

18 : 32

9 : 16