physic help!

(i) Net force acting on the block = ( 80 - 10 - 10 g.sin(30)) N = 20 N
where g is the acceleration due to gravity, taken to be 10 m /s2
Acceleration = 20/10 ms/2 = 2 m /s2
Use equation of motion: s = ut + (1/2).a.t^2
with u = 0 m /2, a = 2 m /s2, s = 10 m , t = ?
hence, 10 = (1/2).(2).t^2
i.e. t = 3.162 s

(ii) Work done by the man = 80 x 10 J = 800 J
Average power = work done/time = 800/3.162 watts = 253 watts

(iii) Use equation of motion: v^2 = u^2 + 2.a.s
with u = 0 m /2, a = 2 m /s2, s = 10 m , v = ?
hence, v^2 = 2 x 2 x 10 (m/s)^2
i.e. v = 6.325 m /s
Gain in kinetic energy = (1/2) x 10 x 40 J = 200 J

(iv) Let v be the terminal speed,
use power = froce x velocity
assume the frictional force remains the same,i.e. 10 N
hence, 10v = 253


2010-05-15 23:58:39 補充：
i.e. v = 253/10 m/s 25.3 m/s

Answer:
(i)3.16s
(ii)253W
(iii)200J
(iv)25.3ms^-1

i. Along the plane, net force acting on the block, F

= 80 - 10(10)sin30* - 10

= 20 N

By Newton's 2nd law of motion, F = ma

20 = 10a

Acceleration, a = 2 ms^-2

By s = ut + 1/2 at^2

10 = 0 + 1/2 (2)t^2

Time, t = 3.16 s


ii. Gain in G.P.E. = mgh = (10)(10)(10sin30*) = 500 J

Average power, P = E/t = 500/3.16 = 158 W


iii. By v^2 = u^2 + 2as

v^2 = 0 + 2(2)(10)

v = 6.32 ms^-1

Gain in K.E. = 1/2 mv^2 = 1/2 (10)(6.32)^2 = 200 J


iv. Assume the man exerts the same force on the block

By P = Fv

158 = 80v

Terminal speed, v = 1.98 ms^-1