A body of mass 0.1kg released from a height
of 50m reaches the ground with a velocity of
10ms-1. The work done against air resistance
is (g = 10 m/s2)
1) 5J 2) 50J 3) 0 4) 45J
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A body of mass 0.1kg released from a height
of 50m reaches the ground with a velocity of
10ms-1. The work done against air resistance
is (g = 10 m/s2)
1) 5J 2) 50J 3) 0 4) 45J
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Answer:the answer would be the 4th option(45j)
Explanation:I have attached a solution.it is based upon law of conservation of energy
Answer:
OPTION (4) [tex]45 \ J[/tex]
Explanation:
We have been given that:
We have to find out the work done against air resistance.
Solution:
We know that the [tex]Potential \ energy = mgh[/tex]
⇒ [tex]0.1 \times 10 \times 50[/tex]
⇒ [tex]50 \ J[/tex]
We know that the [tex]Kinetic \ energy = \frac{1}{2} mv^{2}[/tex]
⇒ [tex]\frac{1}{2} \times 0.1 \times 10 \times 10[/tex]
⇒ [tex]\frac{1}{2} \times 10[/tex]
⇒ [tex]5 \ J[/tex]
Now, the work done will be equal to [tex]P.E - K.E[/tex]
⇒ [tex]50 \ J - 5 \ J[/tex]
⇒ [tex]45 \ J[/tex]
Final Answer:
A body of mass [tex]0.1 \ kg[/tex] released from a height of [tex]50 \ m[/tex] reaches the ground with a velocity of [tex]10 ms^{-1}[/tex]. The work done against air resistance is [tex]45 \ J[/tex].
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