The acceleration of positive performing SHM is 12cm/sec at distance of 3cm from the mean position its time period is?
1 answer:
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Answer:
Because the force one needs to apply is reduced.
Explanation:
- Work can be expressed in terms of force and distance as follows,
Work = Force × distance
- To be more precise, the selection of these two parameters should be in a manner that the distance is what the line of action of the force traces.
- It is obvious that for a given amount of work when one parameter is changed the other changes to counteract the change of the first one.
- Imagine the situation shown in the attached figure. The amount of work that needs to be done during the lifting of the mass to a height of
vertically is
.
- On the slope, however, one needs to push the object an
distance up to get the object to that
height ultimately.
- The work he does on the slope is
and since
,
.
- The force that needs to be applied is less than that applied in the vertical lift. Hence it feels easier. #SPJ4
Answer: magnitude of applied force is FA = mg + F
Where F is the resultant force downward that the rope moves with
Explanation:
Force downwards F is,
F = FA - T
T is the upwards tension force on the rope
FA is the actual applied force in pulling the rope down.
Therefore, T = FA - F .....equ. (1)
For the box to move up with force ma ( it's mass times its acceleration upwards) upwards tension on the roap must exceed its own weight mg ( it's mass times acceleration due to gravity 9.8m/s^2)
Therefore, ma = T - mg
T = ma + mg ..... equ. (2)
Equating equ. 1 and 2
T = FA - F = ma + mg
Therefore FA = ma + mg + F
But at constant velocity a = 0
Magnitude of applied force becomes
FA = mg + F
See image below
The average speed is the ratio between the total space and the total time of the motion:
The total space is
while the total time is
So, the average velocity is
We can also rewrite it in m/s. The total space is
, while the time is 
, and so
The total space is
while the total time is
So, the average velocity is
We can also rewrite it in m/s. The total space is