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A mass of 2000 kg is currentiy being lowered at a velocity of 2 m/s from the drum as shown in the figure. The mass moment of inertia of the drum is 150 kg-m2. On applying the brake, the mass is brought to rest in a distance of 0.5 m. The energy absorbed by the brake (in kJ) is ________.
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- 14.11 kJ
- 13.11 kJ
- 12.11 kJ
- 15.11 kJ
Correct Option: A
V = rω
| ωi = | = | = 2 rad/sec | r | 1 |
| Loss in KE of drum = | J(ω²i - ω²f) | 2 |
| = | 150(2² - 0²) | 2 |
= 300 Joule
| Loss in KE of block = | m(ω²i - ω²f) | 2 |
| = | 2000(2² - 0²) | 2 |
= 4000 Joule
Loss in potential energy of block = mgh
= 2000 × 9.81 × 0.5 = 9810 Joul
Total loss of energy = 300 + 4000 + 9810
= 14110 J
= 14.11 kJ
