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Question: A 0.1kg yo-yo is initially hanging at rest at the end of its string. Then the tension force in the s...

A 0.1kg yo-yo is initially hanging at rest at the end of its string. Then the tension force in the string is adjusted to pulling upward. Which one of the following describes how the yo-yo will move once the string’s force is adjusted?
(A) The yo-yo will accelerate upwards
(B) The yo-yo will accelerate downwards
(C) The yo-yo moves upward at a constant velocity
(D) The yo-yo move downward at a constant velocity

Explanation

Solution

Hint : When the yo-yo was hanging at rest the tension in the string balances the weight of the yo-yo. Acceleration occurs in the direction of the larger force. So from the equation of motion of the yo-yo, we can find the correct answer.

Formula used: In this solution we will be using the following formula;
FNET=ma{\vec F_{NET}} = m\vec a where a\vec a is the acceleration, mm is the mass of the body and FNET{\vec F_{NET}} is the net force on an object.

Complete step by step answer
In the case of the string hanging at rest, applying the Newton’s second law of motion gives,
Tmg=0T - mg = 0 , (assuming downward as negative and upward as positive), hence,
T=mgT = mg ,
By inserting the known values
T=0.1×10=1NT = 0.1 \times 10 = 1N
However, as mentioned in the question, the tension is adjusted to 0.60.6 Newton. Compared to the first tension, this signifies that the tension force was reduced, hence, by Newton’s second law again, we get that
mgT=mamg - T = ma ( assuming downward to be positive and upward as negative)
By inserting known values, we have that
10.6=0.1a1 - 0.6 = 0.1a
0.1a=0.40.1a = 0.4
Dividing both sides by 0.10.1
a=4 ms1\Rightarrow a = 4{\text{ m}}{{\text{s}}^{ - 1}}
The fact that the acceleration is positive shows that the direction is downward (since we already assumed downward as positive)
Hence, the correct answer is B, the yo-yo will accelerate downward.

Note
To avoid confusion, the convention used is matter of convenience, any direction could be taken as positive or negative as in: assuming downward is negative and upward is positive, we have
Tmg=maT - mg = m\vec a , the vector a\vec a implies that we have assumed unknown direction,
Hence, by inserting values
0.61=0.1a0.6 - 1 = 0.1\vec a
0.4=0.1a\Rightarrow - 0.4 = 0.1\vec a
Thus, dividing by 0.10.1 , we get
a=4ms1\vec a = - 4m{s^{ - 1}} , the negative value implies downward acceleration, since in our convention, downward is negative.
Also, alternatively without calculation, we can follow the reasoning that since the second tension is less than the first tension (which is equal to the weight), then, weight will be greater than the second tension, hence accelerate downwards.