Conservation of mechanical energy – problems and solutions

1. An m-kg block is released from the top of the smooth inclined plane, as shown in the figure below. Comparison between the gravitational potential energy and kinetic energy of the block at point M is…

Solution

Gravitational potential energy at point M :

PE_M = m g (1/3h) = 1/3 m g h

Kinetic energy at point M :

KE_M = EP = m g (2/3 h) = 2/3 m g h

EP_M : EK_M

1/3 m g h : 2/3 m g h

1/3 : 2/3

1 : 2

[irp]

2. An ice skier sliding from a height of A, as shown in the figure below. If the initial velocity = 0 and acceleration due to gravity is 10 ms^-2, then what is the velocity of the skier at point B.

Known :

Initial velocity (v_o) = 0

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

The change in height = 50 meters – 10 meters = 40 meters

Wanted: Velocity of the skier at point B

Solution :

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

The initial mechanical energy = potential energy at point B (height = 40 meters)

The final mechanical energy = kinetic energy at point B

The final mechanical energy = The initial mechanical energy

½ m v_t² = m g h

½ v_t² = g h

½ v_t² = (10)(50-10)

½ v_t² = (10)(40)

½ v_t² = 400

v_t² = (2)(400) = 800

v_t = √800 = √(2)(400) = 20√2 m/s

[irp]

3. An object start to sliding from a point of A without the initial velocity. If there is no friction force, what is the velocity of the object at the lowest point.

Known :

Mass of object = m

Initial velocity (v_o) = 0

Height (h) = 20 meters

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

Wanted : Final velocity (v_t)

Solution :

Initial mechanical energy (ME₁) = Gravitational potential energy at point A (PE_A) = m g h = (m)(10)(20) = 200 m

Final mechanical energy (ME₂) = kinetic energy (KE) = ½ m v_t²

The velocity of the object at the lowest point (v_t) ?

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

EM1 = EM₂

200 m = ½ m v_t²

200 = ½ v_t²

400 = v_t²

v_t = 20 m/s

[irp]

4. A 2-kg ball free fall from point A, as shown in figure below (g = 10 ms^-2). After arrive at point B, the kinetic energy = 2 times the potential energy. What is the height of point B above the surface of earth.

Known :

Mass of ball (m) = 2 kg

Acceleration due to gravity (g) = 10 ms^-2

Height of point A (h_A) = 90 meters

Wanted: Height of point B (h_B)

Solution :

When arriving at point B, the kinetic energy of ball at point B = 2 times gravitational potential energy at point B.

EK = 2 EP

½ m v² = 2 m g h_B

½ v² = 2 g h_B

v² = 2(2)(10) h_B

v² = 40 h_B

Velocity (v) of ball when arrive at point B after free fall from point A :

v² = 2 g h = 2(10)(90–h_B) = 20(90–h_B)

Substitute v² at above equation with v² at this equation.

v² = 40 h_B

20(90–h_B) = 40 h_B

1800–20 h_B = 40 h_B

1800 = 40 h_B + 20 h_B

1800 = 60 h_B

h_B = 1800 / 60

h_B = 30 meters

[irp]

5. A 1-kg ball is released and slides down from point A to point C, as shown in figure below. If the acceleration due to gravity = 10 m.s^-2, what is the kinetic energy of ball when arrive at point C.

Known :

Mass of ball (m) = 1 kg

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

The change in height (h) = 0.75 m

The initial potential energy (PE_o) = m g h = (1)(10)(0.75) = 7.5 Joule

Initial kinetic energy (KE_o) = ½ m v_o² = ½ m (0)² = 0

The final gravitational potential energy (PE_t) = m g h = m g (0) = 0

Wanted : Kinetic energy of ball at point C

Solution :

The initial mechanical energy (ME_o) = The final mechanical energy (ME_t)

The initial gravitational potential energy (PE_o) + the initial kinetic energy (KE_o) = The final gravitational potential energy (PE_t) + the final kinetic energy (KE_t)

7.5 Joule + 0 = 0 + KE_t

7.5 Joule = KE_t

Kinetic energy of ball at point C = 7.5 Joule.

6. A 2-kg ball is released from point A, as shown in figure below. The curve plane is smooth. If the acceleration due to gravity is 10 m.s^-2, what is the kinetic energy of ball at point B.

Known :

Mass of ball (m) = 2 kg

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

The change in height (h) = 120 cm = 1.2 meters

The initial gravitational potential energy (PE_o) = m g h = (2)(10)(1.2) = 24 Joule

The initial kinetic energy (KE_o) = ½ m v_o² = ½ m (0)² = 0

The final gravitational potential energy (PE_t) = m g h = m g (0) = 0

Wanted : Kinetic energy of ball at point B

Solution :

The mechanical energy at point A (initial mechanical energy) = the mechanical energy at point B (the final mechanical energy)

The initial mechanical energy (ME_o) = the final mechanical energy (ME_t)

The initial gravitational potential energy (PE_o) + the initial kinetic energy (KE_o) = the final gravitational potential energy (PE_t) + the final kinetic energy (KE_t)

24 Joule + 0 = 0 + KE_t

24 Joule = KE_t

Kinetic energy (KE) of ball at point C = 24 Joule.

7. An object has a fixed mechanical energy, greater kinetic energy, and a smaller gravitational potential energy. The object is…

A. At rest

B. move upward

C. move downward

D. Accelerated upward

Solution :

Kinetic energy is getting bigger means the velocity of the object is getting bigger and the gravitational potential energy is smaller means that the height of the object from the soil surface is smaller. This happens when the object moves from a certain height down to the ground.

The relationship between speed and kinetic energy is shown by the kinetic energy formula:

KE = ½ m v²

Description of formula: KE = kinetic energy, m = mass, v = speed

The relationship between height and gravitational potential energy is represented by the gravitational potential energy formula:

PE = m g h

Description of formula: PE = gravitational potential energy, m = mass, g = acceleration of gravity, h = height

The correct answer is C.