Mechanical energy – problems and solutions

The work-mechanical energy principle

1. The coefficient of the kinetic friction between block and floor (μk) is 0.5. What is the displacement of an object (s)? Acceleration due to gravity is 10 m/s2.

Known :Mechanical energy – problems and solutions 1

The coefficient of the kinetic friction (μk) = 0.5

Mass of block (m) = 4 kg

Acceleration due to gravity (g) is 10 m/s2

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 (v1) = 5 m/s

Final velocity (v2) = 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

fk d = ME2 – ME1

Work done by the kinetic friction force :

W = fk d = (μk)(N)(d) = (0.5)(40)(d) = 20 d

The change of the mechanical energy :

ΔME = ME2 – ME1 = (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 (ΔPE = PE2 – PE1 = 0). Thus the change of the mechanical energy just involves the change of the kinetic energy.

ΔME = KE2 – KE1 = ½ m v22 – ½ m v12 = ½ m (v22 – v12)

ΔME = ½ (4)(02 – 52) = (2)(25) = 50

Displacement of block :

W = ΔME

20 s = 50

s = 50 / 20

s = 2.5 meters

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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 :

v2 = 2 g (2h) = 4 g h

The kinetic energy of object B :

KEB = ½ m v2 = 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 v2.

The principle of conservation of mechanical energy :

m g h = ½ m v2

Because m g h = ½ m v2 then we can change m g h in equation 1 with ½ m v2.

The kinetic energy of object B = 2 m g h = 2(½ m v2) = m v2

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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 :Mechanical energy – problems and solutions 2

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.

Mechanical energy – problems and solutions 3

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 metersMechanical energy – problems and solutions 4

The final mechanical energy = the kinetic energy = 1/2 m v2

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

MEo = MEt

50 m = 1/2 m v2

50 = 1/2 v2

2 (50) = v2

100 = v2

v = 10 m/s

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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 :Mechanical energy – problems and solutions 5

PEM = 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

KEM = PE = m g (0.7 h)

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

PEM : KEM

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

0.3 : 0.7

3 : 7

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6. If PEQ and KEQ have the potential energy and the kinetic energy at point Q (g = 10 m/s2), then PEQ : KEQ =…

Solution :

The gravitational potential energy at point Q :Mechanical energy – problems and solutions 6

PEQ = 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

KEQ = PE = m g h = m (10)(3.2) = 32 m

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

PEQ : KEQ

18 m : 32 m

18 : 32

9 : 16