Question

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​

Answer 1

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 2

Answer:

OPTION (4) $45 \ J$

Explanation:

We have been given that:

• mass of the body = $0.1 \ kg$
• height = $50 \ m$
• velocity = $10ms^{-1}$
• acceleration due to gravity = $10ms^{2}$

We have to find out the work done against air resistance.

Solution:

We know that the $Potential \ energy = mgh$

⇒ $0.1 \times 10 \times 50$

⇒ $50 \ J$

We know that the $Kinetic \ energy = \frac{1}{2} mv^{2}$

⇒ $\frac{1}{2} \times 0.1 \times 10 \times 10$

⇒ $\frac{1}{2} \times 10$

⇒ $5 \ J$

Now, the work done will be equal to $P.E - K.E$

⇒ $50 \ J - 5 \ J$

⇒ $45 \ J$

Final Answer:

A body of mass $0.1 \ kg$ released from a height of $50 \ m$ reaches the ground with a velocity of $10 ms^{-1}$. The work done against air resistance is $45 \ J$.

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