十萬火急 MECHANIC - KINETICS (動力學)

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2011-04-14 4:32 am

識幾多答幾多啦，感激萬分! ^^＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊Useful equations of uniformly accelerated motions (Linear) s = 1/2 (u + v) t v = u + a ts = u t + 1/ 2 a t²v2 = u2 + 2 a s s = Displacement / u = Initial Velocity / v = Final Velocity / a = Acceleration / t = Time＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊＊ 1) A body of mass 10 kg is moved from rest by a force of 100 N. What is the acceleration of the body?

2) A force of 500 N acts on an object to move it 10 m in 2 second. Find the power required.

3) A 1kg object is fired vertically upwards from the ground with a speed of 100 m /s.
a) Find the maximum it can rise.
b) Its' kinetic and potential energy after 2 seconds.

4) A hammer of mass 4 kg , and moving at 6 m/s, strikes a body of mass 12 kg at rest and both hammer and the body move on freely together after the impact, find the gain of kinetic energy of the body after the impact.

5) A 747 jet plane draws 4500 kg /s into its engine and ejects it at a speed of 360 m /s relative to the engine. Find the thrust exerted by the jet, (i) when stationary. (ii) at a forward speed of 600 km/h .

6) Three forces, A, B, and C, act in the same vertical plane from a point O. Force A is 30 N and acts horizontally to the left of O. Force B is 40 N and acts vertically upwards. Determine with the aid of a force diagram the value of force C and the direction in which it acts, if the forces are in equilibrium.

7) Three cords, A, B, and C, are attached to one another at a point O. The angle between cords A and B is 80° and that between B and C is 150°. If the force exerted by cord A is 100 N, determine graphically the values of the forces exerted by cords B and C, assuming the three forces to be in equilibrium.

回答 (1)

Godfrey

2011-04-14 4:41 pm

✔ 最佳答案

1)
Force = mass x acceleration
100 N = 10 kg x acceleration
acceleration = 100/10 = 10 m/s^2

2)
Power = work done/time
Power = Force x distance/time
Power = (500N x 10 m)/2s = 2500 W

3)
a) Find the maximum it can rise.
u = 100 m/s (initial velocity)
v = 0 m/s (final velocity)
a = -9.8 m/s^2 (acceleration due to gravity)
s = ? (distance traveled)
v^2 = u^2 + 2as
0 = 100^2 –(2)(9.8)s
s = 10000/19.6 = 510.2 m
Maximum height it can rise to = 510.2 m

b) Its kinetic and potential energy after 2 seconds
After 2 s, the velocity, v
a = (v – u)/t
.-9.8 = (v – 100)/2
-19.6 = v – 100
v = 100 – 19.6 = 80.4 m/s
K.E. = ½ mv^2
K.E. = ½(1)(80.4)^2 = 3232 J
Kinetic energy = 3232 J

s = ut + 1/2at^2
s = 100(2) -1/2(9.8)(2)^2
s = 100(2) -1/2(9.8)(2)^2
s = 180.4 m
P.E. = mgh = (1)(9.8)(180.4) = 1768 J (Energy is a scalar quantity)
Potential energy = 1768 J

4)
K.E. of hammer = ½ mv^2 = ½(4)(6)^2 = 18 J
After impact, total K.E. of hammer and body is still 18 J
18 J = ½(4 + 12)v^2 = 8v^2, v = Square root of (18/8)
Velocity of hammer and body after the impact, v = 1.5 m/s
K.E. of body = = ½ mv^2 =½ (12)(1.5)^2 = 13.5 J
The gain of K.E. of the body = 13.5 J

I have to go now. Let the other people do the rest. A few hours ago, I also did 3 questions on another page for you. Good luck!

2011-04-15 07:26:12 補充：
5) Not so sure about (b)
(a) thrust = (4500 kg/s) x (360m/s) = 1620000 N
(b) forward speed in m/s = 600 x 1000/3600 =166.66 m/s
Thrust = 4500 kg/s x (360m/s – 166.66 m/s) = 870030 N

6) C = 50 N at S 36.9 deg E
7) B = 153 N, C = 197 N

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