Question

On reaching the top of a slope 6 m high from its bottom, a cyclist has a speed of 1.5 ms -1. Find the kinetic energy and the potential energy of the cyclist. The mass of the cyclist and his bicycle is 40 kg.

Answer 1

Given:-

→ Velocity of the cyclist = 1.5m/s

→ Height of the slope = 6m

→ Mass of the cyclist and his

bicycle = 40kg

To find:-

→ K.E. of the cyclist

→ P.E. of the cyclist

Solution:-

• Acceleration due to gravity = 9.8m/s²

We know that :-

K.E. = 1/2mv²

Where:-

• K.E. is kinetic energy of the body

• m is mass of the body

• v is velocity of the body

=> K.E. = 1/2mv²

=> K.E. = 1/2×40×1.5×1.5

=> K.E. = 20×2.25

=> K.E. = 45 J

Now:-

P.E. = mgh

Where:-

• P.E. is potential energy of the body.

• m is mass of the body.

• g is acceleration due to gravity.

• h is height.

=> P.E. = 40×9.8×6

=> P.E. = 2352 J

Thus:-

• Kinetic energy of the cyclist is 45 J

• Potential energy of the cyclist is 2352 J

Answer 2

Given,

• Height (h) = 6 m
• Mass (m) = 40 kg
• Velocity (v) = 1.5 m/s
• Acceleration due to gravity (g) = 9.8 m/s²

To Find,

• The kinetic energy and the potential energy of the cyclist.

According to the question,

First we ask to find kinetic energy of cyclist.

★ K.E = ¹/2 × mv²

[ Put the values ]

⇢ K.E = ¹/2 × 40 × (1.5)²

⇢ K.E = 20 × 2.25

⇢ K.E = 45 J ★

Now we ask to find Potential energy of cyclist,

★ P.E = mgh

[ Put the values ]

⇢ P.E = 40 × 9.8 × 6

⇢ P.E = 392 × 6

⇢ P.E = 2352 J ★