/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 33 A block of \(10 \mathrm{~kg}\) i... [FREE SOLUTION] | 91Ó°ÊÓ

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A block of \(10 \mathrm{~kg}\) is pulled by a constant speed on a rough horizontal surface by a force of \(19.6 \mathrm{~N}\). The co-efficient of friction is (A) \(0.1\) (B) \(0.2\) (C) \(0.3\) (D) \(0.4\)

Short Answer

Expert verified
The coefficient of friction is (B) \(0.2\).

Step by step solution

01

Identify the forces acting on the block

The forces acting on the block are the gravitational force (weight) acting vertically downwards, the normal force acting vertically upwards, the applied force acting horizontally, and the frictional force acting horizontally opposite to the applied force.
02

Calculate the weight of the block

The weight of the block is given by the product of its mass and the acceleration due to gravity: W = mg where W is the weight, m is the mass (10 kg), and g is the acceleration due to gravity (9.8 m/s²). Substitute the given values to get: W = 10 × 9.8 W = 98 N
03

Calculate the normal force acting on the block

Since the block is moving horizontally, the normal force is equal to the weight of the block (i.e., N = W = 98 N).
04

Calculate the frictional force acting on the block

As the block is moving at a constant speed, the net force acting on it is zero. Therefore, the frictional force (f) is equal to the applied force (F) but in the opposite direction: f = F = 19.6 N.
05

Calculate the coefficient of friction

The frictional force can be calculated using the formula: f = μN where f is the frictional force, μ is the coefficient of friction, and N is the normal force. We can rearrange this equation to solve for μ: μ = f/N Now, we can substitute the values for f and N to find the coefficient of friction: μ = 19.6/98 μ = 0.2 The correct answer is (B) \(0.2\).

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Most popular questions from this chapter

A block \(P\) of mass \(4 \mathrm{~kg}\) is placed on horizontal rough surface with co-efficient of friction \(\mu=0.6\). And two blocks \(R\) and \(Q\) of masses \(2 \mathrm{~kg}\) and \(4 \mathrm{~kg}\) connected with the help of massless strings \(A\) and \(B\), respectively, passing over frictionless pulleys as shown, then \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\) (A) Acceleration of block \(P\) is zero. (B) Tension in string \(A\) is \(20 \mathrm{~N}\). (C) Tension in string \(B\) is \(40 \mathrm{~N}\). (D) Contact force on block \(P\) is \(20 \sqrt{5} \mathrm{~N}\).

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A metre scale is moving with uniform velocity. This implies (A) the force acting on the scale is zero, but a torque about the centre of mass can act on the scale. (B) the force acting on the scale is zero and the torque acting about centre of mass of the scale is also zero. (C) the total force acting on it need not be zero but the torque on it is zero. (D) neither the force nor the torque needs to be zero.

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A string of length \(L\) and mass \(M\) are lying on a horizontal table. A force \(F\) is applied at one of its ends. Tension in the string at a distance \(x\) from the ends at which force is applied is (A) Zero (B) \(F\) (C) \(F(L-x) / L\) (D) \(F(L-x) / M\)

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