Physics Question on Work,Energy and Power

(a) u=0 m/s, s= -10 m, a = g = -10 m/s2, v =?

using v^2 =u^2 + 2.a.s v^2 = 2.(-10)(-10) m2/s2 v = 14.14 m.s

(b) Let F be the force given by water,

using work done by F = decrease of kinetic energy + decrease of potential energy

F.(4) = (1/2).(58)(200) + (58).(10).(4)

solve for F gives F = 2030 N

a. Take the accleration due to gravity, g = 10 ms^-2
Take the direction downward as positive.

Assume no energy is lost to the surroundings.

By the law of conservation of energy

P.E. lost = K.E. gained
mgh = 1/2 mv^2
2gh = v^2
v^2 = 2gh
v^2 = 2(10)(10)
v^2 = 200
v = 14.1 ms^-1

Therefore, the speed of Mr. Hu just before he reaches the water surface is 14.1 ms^-1.


b. By the law of conservation of energy
K.E. lost = P.E. gained + work done against resistive force
1/2 mv^2 = mgh + Fh
1/2 (58)(200) = (58)(10)(4) + F(4)
4F = 3480
F = 870 N

So, the magnitude of the average force exerted on Mr. Hu by water is 870 N.

A) PE lost = KE gain
58*10*10 = 1/2*58*v^2
v =開方200^-2 or 14.1ms^-2 (3 sig fig)
B) work done by the force = Total energy loss
Fs= KE - PE
4F = 1/2開方2^2*58 - 58*10*(-4)
F = 594.5N
The magnitude of the average force exerted on him by the water is 594.5N

2007-04-17 22:53:29 補充：
樓上唔記得左個height 係-4m 所以b)應該計錯