Question

.. The frictional force of the air on a body of mass
0.25 kg, falling with an accleration of 9.2 m/s2,
will be :
(1) 1.0 N
(2) 0.55 N
(3) 0.25 N
145 0.15 N​

Answer 1

Answer:

• The Frictional Force (F) acting on the block is 0.15 N

Given:

1. Mass of body (M) = 0.25 Kg.
2. Acceleration of body (a) = 9.2 m/s²
3. Acceleration due to gravity (g) = 9.8 m/s²

Explanation:

________________________

#Refer the attachment for figure.

From the figure we can make out that Frictional Force (F) acts opposite to the weight of the body (W). As the body is falling downwards the weight of the body is greater than the frictional force acting i.e. W > F

Now, applying Newton's Second law of motion.

⇒ F_{net} = M a

• F_{net} Denotes the Net Force acting on body.
• M Denotes Mass.
• a Denotes acceleration.

Now,

⇒ F_{net} = M a

As the body is falling downwards,

⇒ W - F = M a

⇒ M g - F = M a    ∵[W = M g]

⇒ M g - M a  = F

⇒ F = M g - M a

Substituting the values,

⇒ F = 0.25 × 9.8 - 0.25 × 9.2

⇒ F = 0.25 × (9.8 - 9.2)

⇒ F = 0.25 × 0.6

⇒ F = 0.15

⇒ F = 0.15 N

∴ The Frictional Force (F) acting on the block is 0.15 N.

Hence, Option-4 is correct !

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Answer 2

$\huge\underline\mathbb\red{ANSWER:-}$

0.15n