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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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gravitational Potential Energy
When an object is dropped from a height, it possesses gravitational potential energy. This type of energy relates to the position of an object in a gravitational field. For the knife edge in the problem, the gravitational potential energy when it is at height \(h\) is calculated using the formula \(U = mgh\). Here:
  • \(m\) stands for the mass of the object (knife edge),
  • \(g\) is the acceleration due to gravity, approximately \(9.8 \, \text{m/s}^2\),
  • \(h\) is the height from which the object is dropped.
Gravitational potential energy is a stored energy, waiting to do work. As the knife descends, this energy converts into kinetic and eventually does work on penetrating the wooden floor.
Work-Energy Principle
The Work-Energy Principle is a fundamental concept in physics. It states that the work done on an object is equal to the change in its kinetic energy. In our problem, this principle helps us calculate the energy transfer when the knife edge penetrates the wood.

Initially, the knife possesses gravitational potential energy as it's at a height. When it falls, this energy turns into kinetic energy and ultimately does work against the resistive force of the wood. The work done is given by:
  • \(W = F_\text{avg} \times d\)
Here, \(F_\text{avg}\) is the average resistive force offered by the wood and \(d\) is the penetration depth. Using the conservation of energy, the potential energy converts fully into work done:
  • \(mgh = F_\text{avg} \times d\)
This equation allows us to solve for the resistive force.
Resistive Force
Resistive force, in this context, refers to the resistance offered by the wood against the penetration of the knife. When an object penetrates a surface, it must overcome this force. Our knife edge problem showcases how to calculate this force by equating it with the loss of gravitational potential energy.

To find the resistive force, we rearrange the equation from the Work-Energy Principle:
  • \(F_\text{avg} = \frac{mgh}{d}\)
By factoring out gravity, we derive:
  • \(F_\text{avg} = mg \times \frac{h}{d}\)
This expression tells us how the force is dependent on the initial height \(h\) (where more height means more force), the mass \(m\), and the penetration depth \(d\). The deeper the knife penetrates, the lower the average resistive force.
Physics Problem Solving
Solving physics problems effectively requires a structured approach. The key is to break down the problem using fundamental principles, like the Conservation of Energy, and systematically work through it.

For our exercise, the steps included:
  • Identifying the gravitational potential energy at the start.
  • Applying the Work-Energy Principle to set up the energy balance.
  • Deriving the resistive force by re-arranging the energy equation.
  • Comparing the derived solution with given options to select the correct answer.
This logical breakdown not only helps in arriving at the right answer but also increases your understanding and application of physical laws in real-world scenarios. Each step clarifies the physical interactions that control how objects behave under different forces.

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Most popular questions from this chapter

When \(2 \mathrm{~kg}\) mass hangs to a spring of length \(50 \mathrm{~cm}\), the spring stretches by \(2 \mathrm{~cm}\). The mass is pulled down until the length of the spring becomes \(60 \mathrm{~cm}\). What is the amount of elastic energy stored in the spring in this condition, if \(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\) (A) \(10 \mathrm{~J}\) (B) \(2 \mathrm{~J}\) (C) \(2.5 \mathrm{~J}\) (D) \(5 \mathrm{~J}\)

A metal ball of mass \(2 \mathrm{~kg}\) moving with a velocity of \(36 \mathrm{~km} / \mathrm{h}\) has a head on collision with a stationary ball of mass \(3 \mathrm{~kg}\). If after the collision, the two balls move together, the loss in kinetic energy due to collision is (A) \(40 \mathrm{~J}\) (B) \(60 \mathrm{~J}\) (C) \(100 \mathrm{~J}\) (D) \(140 \mathrm{~J}\)

Two bodies \(\mathrm{P}\) and \(\mathrm{Q}\) have masses \(5 \mathrm{~kg}\) and \(20 \mathrm{~kg}\) respectively. Each one is acted upon by a force of \(4 \mathrm{~N}\). If they acquire the same kinetic energy in times \(t_{\mathrm{P}}\) and \(\mathrm{t}_{\mathrm{Q}}\) then the ratio \(\left(t_{q} / t_{p}\right)=\ldots \ldots\) \((\mathrm{A})(1 / 2)\) (B) 2 (C) 5 (D) 6

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