Question

The mass of a cyclist together with the
bicycle is 90 kg. Calculate the work done
by cyclist if the speed increases from
6km/h to 12 km/h.​

Answer 1

Given:-

→ Mass of the cylist with bicycle = 90 kg

→ Initial velocity of the cyclist = 6 km/h

→ Final velocity of the cyclist = 12 km/h

To find:-

→ Work done by the cyclist.

Solution:-

Firstly let's convert the initial and final velocities of the cyclist from km/h to m/s.

Initial velocity:-

⇒ 1 km/h = 5/18 m/s

⇒ 6 km/h = 6×5/18

⇒ 5/3 m/s

Final velocity:-

⇒ 12 km/h = 12×5/18

⇒ 10/3 m/s

________________________________

Now, we know that :-

Work done = Change in K.E.

⇒ W = 1/2mv² - 1/2mu²

⇒ W = m/2[v² - u²]

⇒ W = 90/2[(10/3)² - (5/3)²]

⇒ W = 45[100/9 - 25/9]

⇒ W = 45[75/9]

⇒ W = 5 × 75

⇒ W = 375 J

Thus, required work done is 375 J .

Answer 2

Given :-

• Mass = 90 kg
• Initial velocity = 6 km/h
• Final velocity = 12 km/h

To Find :-

Work done

Solution :-

Here,

At first converting velocity into m/s

6 × 5/18 = 30/18 = 5/3 m/s = 1.667 m/s

12 × 5/18 = 60/18 = 10/3 = 3.33 m/s

Now

We are knowing that

The change in KE is the difference between the initial KE and final KE.

So,

W = 1/2mv² - 1/2mu²

Taking 1/2 and m as common

W = m/2[v² - u²]

W = 90/2[3.33² - 1.67²]

W = 90/2[11.088 - 2.788]

W = 45[8.3]

W = 373.5 J