Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 440

An open knife edge of mass \(\mathrm{m}\) is dropped from a height \(\mathrm{h}\) on a wooden floor. If the blade penetrates up to the depth d into the wood, the average resistance offered by the wood to the knife edge is, (A) \(\mathrm{mg}\) (B) \(\mathrm{mg}(1+\\{\mathrm{h} / \mathrm{d}\\})\) (C) \(\mathrm{mg}(1+\\{\mathrm{h} / \mathrm{d}\\})^{2}\) (D) \(m g(1-\\{h / d\\})\)

Short Answer

Expert verified
The short answer to the question is: The average resistance offered by the wood to the knife edge is (B) \(\mathrm{mg}(1+\\{\mathrm{h} / \mathrm{d}\\})\).

Step by step solution

01

Define initial and final energy

The initial energy is the gravitational potential energy, which is given by \(U = mgh\). The final energy is the work done against the resistive force, given by \(W = F_\text{avg} \times d\). Thus, using the conservation of energy, we get: \[mgh = F_\text{avg} \times d\]
02

Solve for the average force

Solving the equation above for the average force \(F_\text{avg}\), we get: \[F_\text{avg} = \frac{mgh}{d}\]
03

Factor out gravity

To match the answer choices with the parameters given in the problem, we will factor out the gravitational acceleration "g" like this: \[F_\text{avg} = mg \times \frac{h}{d}\]
04

Identify the answer choice

Now, looking at the given answer choices, we see that our derived equation matches with option (B): \(mg(1+\\{h / d\\})\) So the average resistance offered by the wood to the knife edge is (B) \(\mathrm{mg}(1+\\{\mathrm{h} / \mathrm{d}\\})\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A body of mass \(6 \mathrm{~kg}\) is under a force, which causes a displacement in it given by $\mathrm{S}=\left[\left(2 \mathrm{t}^{3}\right) / 3\right](\mathrm{in} \mathrm{m}) .$ Find the work done by the force in first one seconds. (A) \(2 \mathrm{~J}\) (B) \(3.8 \mathrm{~J}\) (C) \(5.2 \mathrm{~J}\) (D) \(24 \mathrm{~J}\)

Two bodies of masses \(\mathrm{m}\) and \(3 \mathrm{~m}\) have same momentum. their respective kinetic energies \(E_{1}\) and \(E_{2}\) are in the ratio..... (A) \(1: 3\) (B) \(3: 1\) (C) \(1: 3\) (D) \(1: 6\)

A body falls freely under the action of gravity from a height \(\mathrm{h}\) above the ground. $$ \begin{array}{|l|l|} \hline \text { Column - i } & \text { Column - ii } \\ \hline \text { (a) P.E. }=2 \text { (K.E.) } & \text { (P) constant at every point } \\ \hline \text { (b) P.E. }=\text { K.E. } & \text { (Q) at height }(\mathrm{h} / 3) \\ \hline \text { (c) P.E. }=(1 / 2) \text { (K.E.) } & \text { (R) at height }(2 \mathrm{~h} / 3) \\ \hline \text { (d) P.E.+ K.E. } & \text { (S) at height }(\mathrm{h} / 2) \\ \hline \end{array} $$ (A) \(a-P, b-Q, c-R, d-S\) (B) $\mathrm{a}-\mathrm{Q}, \mathrm{b}-\mathrm{P}, \mathrm{c}-\mathrm{S}, \mathrm{d}-\mathrm{R}$ (C) $\mathrm{a}-\mathrm{S}, \mathrm{b}-\mathrm{R}, \mathrm{c}-\mathrm{Q}, \mathrm{d}-\mathrm{P}$ (D) $\mathrm{a}-\mathrm{R}, \mathrm{b}-\mathrm{S}, \mathrm{c}-\mathrm{Q}, \mathrm{d}-\mathrm{P}$

Assertion and Reason are given in following questions. Each question have four option. One of them is correct it. (1) If both assertion and reason and the reason is the correct explanation of the Assertion. (2) If both assertion and reason are true but reason is not the correct explanation of the assertion. (3) If the assertion is true but reason is false. (4) If the assertion and reason both are false. Assertion: In the elastic collision between two bodies, the relative speed of the bodies after collision is equal to the relative speed before the collision. Reason: In the elastic collision the linear momentum of the system is conserved. (A)1 (B) 2 (C) 3 (D) 4

A bomb of \(12 \mathrm{~kg}\) divides in two parts whose ratio of masses is $1: 4 .\( If kinetic energy of smaller part is \)288 \mathrm{~J}$, then momentum of bigger part in \(\mathrm{kgm} / \mathrm{sec}\) will be (A) 38 (B) 72 (C) 108 (D) Data is incomplete
See all solutions

Recommended explanations on English Textbooks

Cues and Conventions

Read Explanation

Language Analysis

Read Explanation

English Language Study

Read Explanation

Summary Text

Read Explanation

Single Paragraph Essay

Read Explanation

Prosody

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free