Inactive Tutor answered 03/26/13
work = change of mechanical energy
work
= (1/2)mV22 - [(1/2)mV12 + mgh1]
= (1/2)(77)(8.2)2 - [(1/2)(77)(1.32) + (77)(9.8)(1.65)]
= 1278.585 ~ 1300 J
Attn: The final answer contains only 2 siginificant figures.
Anna R.
asked 03/26/13Catching a wave, 77-kg surfer starts with a speed of 1.3 m/s, drops a height of 1.65m, and ends with a speed of 8.2 m/s. How much non conservative work was done on the suffer?
Inactive Tutor answered 03/26/13
work = change of mechanical energy
work
= (1/2)mV22 - [(1/2)mV12 + mgh1]
= (1/2)(77)(8.2)2 - [(1/2)(77)(1.32) + (77)(9.8)(1.65)]
= 1278.585 ~ 1300 J
Attn: The final answer contains only 2 siginificant figures.
Inactive Tutor answered 03/26/13
The initial energy (E in) of the surfer includes its initial kinetic energy plus potential energy for the height h =1.65 m. Let's put
v1 = 1.3 m/s (initial speed), m = 77 kg, g = 9.81 m/s^2 (acceleration due to gravity).
Then E in = mv1^2/2 + mgh = 1311.43 J (rounded to the nearest hundreds).
Its final energy is equal to its kinetic eenrgy with the speed v2 = 8.2 m/s. The energy equals
mv2^2/2 = 2588.74 J
The difference between these two equals to the work done on the surfer which includes work of forces f friction between the surfer and the water (viscosity), and push of the wave. The difference is 1277.31 J.
Get a free answer to a quick problem.
Most questions answered within 4 hours.
Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.