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Chapter 3: Problem 347

A block of mass \(4 \mathrm{~kg}\) is placed on a rough horizontal plane. A time dependent force \(\mathrm{F}=\mathrm{Kt}^{2}\) acts on a block, where \(\mathrm{k}=2 \mathrm{~N} / \mathrm{s}^{2}\), co-efficient of friction \(\mu=0.8\). Force of friction between the block and the plane at $\mathrm{t}=2 \mathrm{~S}$ is.... (A) \(32 \mathrm{~N}\) (B) \(4 \mathrm{~N}\) (C) \(2 \mathrm{~N}\) (D) \(8 \mathrm{~N}\)

Short Answer

Expert verified
The force of friction between the block and the plane at \(t=2~\mathrm{s}\) is \(8~\mathrm{N}\). The correct answer is (D) \(8~\mathrm{N}\).

Step by step solution

01

Write down the time-dependent force equation

The time-dependent force is given by \(\mathrm{F}=\mathrm{Kt}^2\), where \(\mathrm{K} = 2~\mathrm{N}/\mathrm{s}^2\). We need to find the force of friction acting on the block at \(t = 2~\mathrm{s}\).
02

Calculate the force at the given time

To find the force acting on the block at \(t = 2~\mathrm{s}\), we'll substitute the given time into the time-dependent force equation: \(\mathrm{F} = 2 \times (2~\mathrm{s})^2\) \(\mathrm{F} = 2 \times 4~\mathrm{N}\) \(\mathrm{F} = 8~\mathrm{N}\) So the force acting on the block at \(t = 2~\mathrm{s}\) is \(8~\mathrm{N}\).
03

Find the normal force

Since the block is on a horizontal plane, there is no vertical acceleration or force. Therefore, the normal force (\(N\)) equals the weight of the block: \(N = mg\) \(N = (4~\mathrm{kg})(9.81~\mathrm{m/s^2})\) \(N = 39.24~\mathrm{N}\)
04

Calculate the maximum force of static friction

Using the coefficient of friction \(\mu = 0.8\), we can find the maximum force of static friction (\(F_{s_{max}}\)) between the block and the plane: \(F_{s_{max}} = \mu N\) \(F_{s_{max}} = (0.8)(39.24~\mathrm{N})\) \(F_{s_{max}} = 31.392~\mathrm{N}\)
05

Compare the force with the force of static friction

We can now compare the force acting on the block at \(t = 2~\mathrm{s}\) (\(8~\mathrm{N}\)) with the maximum force of static friction (\(31.392~\mathrm{N}\)). Since the force is less than the maximum force of static friction, the block doesn't move and the force of friction is equal to the force acting on the block. Therefore, the force of friction at \(t = 2~\mathrm{s}\) is \(8~\mathrm{N}\). The correct answer is (D) \(8~\mathrm{N}\).

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Most popular questions from this chapter

A block B, placed on a horizontal surface is pulled with initial velocity \(\mathrm{V}\). If the coefficient of kinetic friction between surface and block is \(\mu\), than after how much time, block will come to rest? (A) (v/g) (B) \((\mathrm{g} / \mathrm{v})\) (C) \((\mathrm{g} / \mathrm{v})\) (D) \((\mathrm{v} / \mathrm{g})\)

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Assertion and reason are given in following question. Each question have four options. One of them is correct select it. (A) Assertion is true. Reason is true and reason is correct explanation for Assertion. (B) Assertion is true. Reason is true but reason is not the correct explanation of assertion. (C) Assertion is true. Reason is false. (D) Assertion is false. Reason is true. Assertion: A body of mass $1 \mathrm{~kg}\( is moving with an acceleration of \)1 \mathrm{~ms}^{-1}$ The rate of change of its momentum is \(1 \mathrm{~N}\). Reason: The rate of change of momentum of body \(=\) force applied on the body. (A) a (B) \(\mathrm{b}\) (C) (D) \(\mathrm{d}\)

A wagon weighing \(1000 \mathrm{~kg}\) is moving with a velocity $50 \mathrm{~km} \mathrm{~h}^{-1}\( on smooth horizontal rails. A mass of \)250 \mathrm{~kg}$ is dropped into it. The velocity with which it moves now is (A) \(2.5 \mathrm{~km} \mathrm{~h}^{-1}\) (B) \(20 \mathrm{~km} \mathrm{~h}^{-1}\) (C) \(40 \mathrm{~km} \mathrm{~h}^{-1}\) (D) \(50 \mathrm{~km} \mathrm{~h}^{-1}\)
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