Anwers – Kinetic Energy and Gravitational Potential Energy

Answer:

Given:

• Mass (m) = 1200 kg
• Velocity (v) = 25 m/s

Using the formula for kinetic energy:

KE = ½ × m × v²

KE = ½ × 1200 × 25²

KE = 600 × 625

KE = 375,000 J

KE = 375 kJ

The kinetic energy of the car is 375,000 joules or 375 kilojoules.

Answer:

Given:

• Mass (m) = 1.5 kg
• Height (h) = 2 m
• Gravitational field strength (g) = 9.8 N/kg

Using the formula for gravitational potential energy:

GPE = m × g × h

GPE = 1.5 × 9.8 × 2

GPE = 29.4 J

The gravitational potential energy of the book is 29.4 joules.

Answer:

Given:

• Mass (m) = 0.3 kg
• Initial velocity (v) = 12 m/s
• Gravitational field strength (g) = 9.8 N/kg

a) Initial kinetic energy:

KE = ½ × m × v²

KE = ½ × 0.3 × 12²

KE = 0.15 × 144

KE = 21.6 J

b) Maximum height:

Using the principle of energy conservation, at maximum height all kinetic energy is converted to gravitational potential energy:

Initial KE = Final GPE

21.6 = m × g × h

21.6 = 0.3 × 9.8 × h

21.6 = 2.94 × h

h = 21.6 ÷ 2.94

h = 7.35 m

c) Gravitational potential energy at maximum height:

GPE = m × g × h

GPE = 0.3 × 9.8 × 7.35

GPE = 21.6 J

The initial kinetic energy is 21.6 J, the maximum height is 7.35 m, and the gravitational potential energy at maximum height is 21.6 J.

Answer:

Given:

• Mass (m) = 800 kg
• Height (h) = 45 m
• Gravitational field strength (g) = 9.8 N/kg

a) Gravitational potential energy at the top:

GPE = m × g × h

GPE = 800 × 9.8 × 45

GPE = 352,800 J

GPE = 352.8 kJ

b) Speed at the bottom:

Using the principle of energy conservation, if all GPE is converted to KE:

GPE = KE

m × g × h = ½ × m × v²

g × h = ½ × v²

v² = 2 × g × h

v² = 2 × 9.8 × 45

v² = 882

v = √882

v = 29.7 m/s

The gravitational potential energy at the top is 352,800 J, and the speed at the bottom would be 29.7 m/s.

Answer:

Given:

• Mass (m) = 0.2 kg
• Speed at lowest point (v) = 2.5 m/s
• Gravitational field strength (g) = 9.8 N/kg

a) Kinetic energy at lowest point:

KE = ½ × m × v²

KE = ½ × 0.2 × 2.5²

KE = 0.1 × 6.25

KE = 0.625 J

b) Height from which pendulum was released:

Using the principle of energy conservation, the gravitational potential energy at release equals the kinetic energy at the lowest point:

GPE = KE

m × g × h = 0.625

0.2 × 9.8 × h = 0.625

1.96 × h = 0.625

h = 0.625 ÷ 1.96

h = 0.319 m

h ≈ 0.32 m or 32 cm

The kinetic energy at the lowest point is 0.625 J, and the pendulum was released from a height of 0.32 m above its lowest point.

Answer:

Given:

• Mass (m) = 90 kg
• Work done/Energy added (KE) = 4500 J
• Initial velocity = 0 m/s (at rest)

Using the formula for kinetic energy to find final velocity:

KE = ½ × m × v²

4500 = ½ × 90 × v²

4500 = 45 × v²

v² = 4500 ÷ 45

v² = 100

v = 10 m/s

The final speed of the cyclist would be 10 m/s.

Answer:

Given:

• Mass (m) = 5 kg
• Height (h) = 8 m
• Gravitational field strength (g) = 9.8 N/kg

a) Initial gravitational potential energy:

GPE = m × g × h

GPE = 5 × 9.8 × 8

GPE = 392 J

b) Kinetic energy just before hitting the ground:

Using the principle of energy conservation (ignoring air resistance):

Initial GPE = Final KE

KE = 392 J

c) Speed just before hitting the ground:

KE = ½ × m × v²

392 = ½ × 5 × v²

392 = 2.5 × v²

v² = 392 ÷ 2.5

v² = 156.8

v = √156.8

v = 12.5 m/s

The initial GPE is 392 J, the final KE is 392 J, and the speed just before hitting the ground is 12.5 m/s.

Answer:

Given:

• Mass (m) = 60 kg
• Height (h) = 4 m
• Time (t) = 6 s
• Gravitational field strength (g) = 9.8 N/kg

a) Gravitational potential energy gained:

GPE = m × g × h

GPE = 60 × 9.8 × 4

GPE = 2352 J

GPE = 2.352 kJ

b) Power developed:

Power = Energy ÷ Time

Power = 2352 ÷ 6

Power = 392 watts (W)

The gravitational potential energy gained is 2352 J, and the power developed is 392 W.

Answer:

Given:

• Mass (m) = 0.05 kg
• Initial velocity (v) = 15 m/s
• Height (h) = 1.2 m
• Gravitational field strength (g) = 9.8 N/kg

a) Initial kinetic energy of the ball:

KE = ½ × m × v²

KE = ½ × 0.05 × 15²

KE = 0.025 × 225

KE = 5.625 J

b) Kinetic energy just before hitting the ground:

Since the ball is fired horizontally, we need to consider both the original KE and the GPE due to height:

GPE from height = m × g × h = 0.05 × 9.8 × 1.2 = 0.588 J

Total KE just before hitting the ground = Initial KE + GPE

Total KE = 5.625 + 0.588 = 6.213 J

c) Speed of the ball just before hitting the ground:

KE = ½ × m × v²

6.213 = ½ × 0.05 × v²

6.213 = 0.025 × v²

v² = 6.213 ÷ 0.025

v² = 248.52

v = √248.52

v = 15.8 m/s

The initial kinetic energy is 5.625 J, the kinetic energy just before hitting the ground is 6.213 J, and the speed just before hitting the ground is 15.8 m/s.