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Chapter 3: Problem 362

The average force necessary to stop a hammer with 25 NS momentum in $0.04 \mathrm{sec}\( is \)\quad \mathrm{N}$ (A) 625 (B) 125 (C) 50 (D) 25

Short Answer

Expert verified
The average force necessary to stop the hammer is calculated using the formula \(F = \frac{\Delta{P}}{\Delta{t}}\). Plugging in the given values, we get \(F = \frac{25 \, \mathrm{Ns}}{0.04 \, \mathrm{sec}} = 625 \, \mathrm{N}\). The correct answer is (A) 625 N.

Step by step solution

01

Write down the known values

We know the following values from the given exercise: - Momentum (P) = 25 Ns - Time (t) = 0.04 sec
02

Write the formula for average force

The formula for average force (F) can be expressed as the change in momentum (∆P) divided by the time interval (∆t). Mathematically, the formula can be written as: \[F = \frac{\Delta{P}}{\Delta{t}}\]
03

Calculate the average force

Now, plug in the known values into the formula: \[F = \frac{25 \, \mathrm{Ns}}{0.04 \, \mathrm{sec}}\] Divide 25 Ns by 0.04 sec to find the average force: \[F = 625 \, \mathrm{N}\]
04

Choose the correct option

The average force necessary to stop the hammer is 625 N, which corresponds to option (A). Therefore, the correct answer is: (A) 625

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Most popular questions from this chapter

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 \(12 \mathrm{~ms}^{-1}\)

With what acceleration (a) should a box descend so that a block of mass \(\mathrm{M}\) placed in it exerts a force \((\mathrm{Mg} / 4)\) on the floor of the box? (A) \((4 \mathrm{~g} / 3)\) (B) \((3 \mathrm{~g} / 4)\) (C) \(\mathrm{g} / 4\) (D) \(3 \mathrm{~g}\)

A car travelling at a speed of \(30 \mathrm{~km} / \mathrm{h}\) is brought to a halt in 8 meters by applying brakes. If the same car is travelling at $60 \mathrm{~km} / \mathrm{h}$ it can be brought to a halt with the same breaking power in (A) \(8 \mathrm{~m}\) (B) \(16 \mathrm{~m}\) (C) \(24 \mathrm{~m}\) (D) \(32 \mathrm{~m}\)

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) \(-67.5 \mathrm{~N}\)
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