3. Why is the answer C?
4. Could you please explain every option?
5. (3) shouldn’t output power of the machine include also the kineticenergy of the load?
please show your steps clearly for calculation problems thanks!!!!
5. That means the same amount of energy is needed to raise two blocks of the same mass at constant velocities of 3 ms^-1 and 5ms^-1 respectively over the same height? What makes the difference in their velocities then? Thanks!!
- 天同Lv 76 年前最愛解答
1. Toal dispalcement = 2 x (1/2) x 3 x 3^2 m = 27 m
Average velocity = 27/6 s = 4.5 s
Hence, only statement (2) is correct.
2. Let T be the tension in the string
Consider the 2 kg mass: F - T - 2g = 2 x 2
Consider the 4 kg mass: T - 4g = 4 x 2
Solve for F gives F = 72 N
Hence, when T = 0,
The 2 kg mass: 72 - 2g = 2a where a is the acceleration
a = (72 - 2g)/2 m/s^2 = 26 m/s^2
The 4 kg mass: 4g = 4a' where a' is the new acceleration
a' = g = 10 m/s^2 (downward)
It seems that no option is correct.
3. Options A and C seem to be the same.
4. The answer is option D.
This just follows from the Law of Conservation of Energy. The work done by the man must equal to the work done on the load.
5. The load is moving at uniform speed, there is NO change of kinteic energy.
2014-07-27 16:40:07 補充：
Your suppl question:
Clearly, the power needed to raise the mass with higher speed ( 5 m/s) is larger.
In time of 1 s interval, this mass is raised by 5 m, the gain in potential energy (PE)
= 5mg, where m is the mass and g the acceleration due to gravity.
2014-07-27 16:45:31 補充：
But for the other mass with lower speed (3 m/s), gain in PE in 1 s = 3mg
Thus, power required to raise the mass with higher speed is larger.
2014-07-27 16:48:38 補充：
Note that power is the "energy per UNIT TIME". Though the gain in PE by both masses rasied through the same height are the same, the mass with higher speed achieves this in a shorter time. This indicates that a larger power is required.