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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) \(\mathrm{mg}(1+\\{\mathrm{h} / \mathrm{d}\\})\).

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Most popular questions from this chapter

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