in water. Find the work done to haul the anchor and chain to the surface of the water from a depth of 25 ft. Break up the chain into segments of height h. If we have already removed hfeet of chain from the water, then the rest of the chain weighs 3(25 h) pounds. Of course, the anchor has constant weight of 100 pounds.

When it is at 3 2 m above the ground, it has a gravitational potential energy of 180kg ×9.8 m s2 × 3 2 m = 2646J. So its gravitational potential energy went from 0 at ground level to 2646N at 3 2 m. In conclusion, 2646J of work is done on the 180kg weight as you lift it 3 2 m above ground. 2646 joules Since the work done here is raising the

The acceleration of both blocks is the same and is equal to (M 2 g)/ (M 1 +M 2 ), or 4.86 m/s 2. The work done on each block is equal to their final kinetic energy and added together equals 40.71 J, which is the total work done on the system. Yes I'd say this is a satisfactory understanding.

^{A man weighing 50 kg f supports a body of 25 kg f on his head. what is the work done when he moves a dis†an ce of 20m up an incline of 1 in 10. take g=9. Login. Study Materials. A 50 k g man with 20 k g load on his head climbs up 20 steps of 0.25 m height each. The work done by the man on the block during climbing is. Q.} Find F. Use this calculator to determine the force exerted on an object given its mass and acceleration. the mass of the object. m. k i l o g r a m. the acceleration of the object. a. m e t e r / s q u a r e s e c o n d. ^{First of all, the work being done by the gas on the atmosphere (9.975 kJ) is greater than the heat added (1.005 kJ) plus the reduction in internal energy (3.56 kJ), in violation of the first law (conservation of energy). Secondly, adding the net work done (4.565 kJ) to the work done on the atmosphere (9.975 kJ) to come up with something called} CG6Z. 149 497 426 228 23 413 400 51 218

find the work done when a 25 kg weight