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Chapter 3: Problem 361

A body of mass \(0.05 \mathrm{~kg}\) is falling with acceleration $9.4 \mathrm{~ms}^{-2}$. The force exerted by air opposite to motion is \(\mathrm{N}\) \(\left(g=9.8 \mathrm{~ms}^{-2}\right)\) (A) \(0.02\) (B) \(0.20\) (C) \(0.030\) (D) Zero

Short Answer

Expert verified
The force exerted by air opposite to motion is \(0.02 \mathrm{ ~N}\), which corresponds to option (A).

Step by step solution

01

Identify the accelerations (gravitational and net)

The body is falling under the influence of gravity (g = 9.8 m/s²) and is accelerated downward. The net acceleration (a_net) of the body is given as 9.4 m/s². Notice that the net acceleration is less than the gravitational acceleration, this is because of the air resistance acting on the body.
02

Write down Newton's second law equation

According to Newton's second law, the net force acting on an object is equal to the product of its mass and acceleration. Mathematically, it can be represented as: \(F_{net} = m \cdot a_{net}\) where \(F_{net}\) is the net force, m is the mass of the body and \(a_{net}\) is the net acceleration.
03

Calculate the gravitational force acting on the body

As the body is falling downward, we have a gravitational force acting on it which can be calculated as: \(F_g = m \cdot g\) where \(F_g\) is the gravitational force, m is the mass of the body, and g = 9.8 m/s² Substituting the given values, we get: \(F_g = 0.05 \mathrm{~kg} \cdot 9.8 \mathrm{~ms}^{-2} = 0.49 \mathrm{ ~N}\)
04

Calculate the net force acting on the body

Now, use the Newton's second law equation to calculate the net force acting on the body: \(F_{net} = m \cdot a_{net}\) Substituting the given values, we get: \(F_{net} = 0.05 \mathrm{~kg} \cdot 9.4 \mathrm{~ms}^{-2} = 0.47 \mathrm{ ~N}\)
05

Determine the force exerted by air opposite to motion

Now that we have the gravitational force and the net force acting on the body, we can determine the force exerted by air opposite to motion, \(F_{air}\), using the equation: \(F_{air} = F_g - F_{net}\) Substituting the calculated values, we get: \(F_{air} = 0.49 \mathrm{ ~N} - 0.47 \mathrm{ ~N} = 0.02 \mathrm{ ~N}\) So, the force exerted by air opposite to motion is 0.02 N, which corresponds to option (A).

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Most popular questions from this chapter

The linear momentum \(\mathrm{P}\) of a particle varies with the time as follows. \(P=a+b t^{2}\) Where \(a\) and \(b\) are constants. The net force acting on the particle is (A) Proportional to t (B) Proportional to t \(^{2}\) (C) Zero (D) constant

A rope which can withstand a maximum tension of \(400 \mathrm{~N}\) hangs from a tree. If a monkey of mass \(30 \mathrm{~kg}\) climbs on the rope in which of the following cases-will the rope break? (take \(g=10 \mathrm{~ms}^{-}{ }^{2}\) and neglect the mass of rope \()\) (A) When the monkey climbs with constant speed of \(5 \mathrm{~ms}^{-1}\) (B) When the monkey climbs with constant acceleration of \(2 \mathrm{~ms}^{-2}\) (C) When the monkey climbs with constant acceleration of \(5 \mathrm{~ms}^{-2}\) (D) When the monkey climbs with the constant speed of \(12 \mathrm{~ms}^{-1}\)

The minimum force required to start pushing a body up a rough (coefficient of) inclined plane is \(\mathrm{F}_{1}\). While the minimum force needed to prevent it from sliding down is \(\mathrm{F}_{2}\). If the inclined plane makes an angle \(\theta\) from the horizontal. such that \(\tan \theta=2 \mu\) than the ratio \(\left(\mathrm{F}_{1} / \mathrm{F}_{2}\right)\) is (A) 4 (B) 1 (C) 2 (D) 3

The velocity of a body of mass \(20 \mathrm{~kg}\) decrease from $20 \mathrm{~ms}^{-1}\( to \)5 \mathrm{~ms}^{-1}\( in a distance of \)100 \mathrm{~m}$. Force on the body is (A) \(-27.5 \mathrm{~N}\) (B) \(-47.5 \mathrm{~N}\) (C) \(-37.5 \mathrm{~N}\) (D) \(-67.5 \mathrm{~N}\)

Two blocks of mass \(8 \mathrm{~kg}\) and \(4 \mathrm{~kg}\) are connected a heavy string Placed on rough horizontal Plane, The \(4 \mathrm{~kg}\) block is Pulled with a constant force \(\mathrm{F}\). The co-efficient of friction between the blocks and the ground is \(0.5\), what is the value of \(F\), So that the tension in the spring is constant throughout during the motion of the blocks? \(\left(\mathrm{g}=10 \mathrm{~ms}^{-2}\right)\) (A) \(40 \mathrm{~N}\) (B) \(60 \mathrm{~N}\) (C) \(50 \mathrm{~N}\) (D) \(30 \mathrm{~N}\)
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