Question

A body of mass 1 kg thrown upwards with a velocity of 10 m/s comes to rest (momentarily) after moving up 4m. The work done by air drag in this process is (Take g = 10 m/s²)

• A. −20J
• B. –10J
• C. −30J
• D. 0 J

Answer 1

Answer: (B)

–10J

Answer 2

According to work energy theorem

W_{gravitational force} + W_{air drag} = change in KE.

⇒ $\mathrm { mg } \cdot \Delta \overrightarrow { \mathrm { r } }$ + W_{air drag} = change in KE.

⇒ − mgh + W_{air drag} = change in KE

⇒ W_{air drag} = change in KE + mgh = [0 − $\frac { 1 } { 2 }$ mu² ] + mgh =

−50 + 40 = −10 Joules.