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Chapter 3: Problem 64

A constant retarding force of \(50 \mathrm{~N}\) is applied to a body of mass \(20 \mathrm{~kg}\) moving initially with a speed of \(15 \mathrm{~ms}^{-1}\). How long does the body take to stop? (A) \(2 \mathrm{~s}\) (B) \(4 \mathrm{~s}\) (C) \(6 \mathrm{~s}\) (D) \(8 \mathrm{~s}\)

Short Answer

Expert verified
The body takes 6 seconds to stop. The correct answer is (C).

Step by step solution

01

List down the given data

The retarding force: \(F = 50\text{ N}\) The mass of the body: \(m = 20\text{ kg}\) Initial velocity: \(v_0 = 15 \text{m/s}\) Final velocity: \(v_f = 0\text{ m/s}\) (since the body stops)
02

Use Newton's second law to relate force, mass, and acceleration

Newton's second law states that \(F = ma\), where F is the force applied, m is the mass of the body, and a is the acceleration of the body. As the force is a retarding force (acting in the opposite direction of the motion), we should consider the force as negative. So, the equation becomes \(-F = ma\). Now we can plug in the given values to find the acceleration: \(-50\mathrm{~N} = (20\text{ kg})a\)
03

Calculate acceleration

To calculate acceleration, we'll rearrange the equation from step 2 and solve for a: \(a = -\frac{50\text{ N}}{20\text{ kg}}\) \(a = -2.5 \text{m/s}^2\)
04

Use the kinematic equation to relate initial and final velocities, acceleration, and time

The kinematic equation that relates initial velocity, final velocity, acceleration, and time is: \(v_f = v_0 + at\) Plug in the values we found so far: \(0\text{ m/s} = 15\text{ m/s} - (2.5 \text{m/s}^2)t\)
05

Solve for time

Rearrange the equation from step 4 and solve for t: \(t = \frac{(0\text{ m/s}-15\text{ m/s})}{-2.5 \text{m/s}^2}\) \(t = \frac{-15}{-2.5}\) \(t = 6\text{ s}\) So, the body takes 6 seconds to stop. The correct answer is (C).

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Most popular questions from this chapter

An insect crawls up a hemispherical surface very slowly (Fig. \(3.88\) ). The coefficient of friction between the insect and the surface is \(1 / 3 .\) If the line joining the centre of hemispherical surface to the insect makes an angle \(\alpha\) with the vertical, the maximum possible value of \(\alpha\) is given by (A) \(\cot \alpha=3\) (B) \(\tan \alpha=3\) (C) \(\sec \alpha=3\) (D) \(\operatorname{cosec} \alpha=3\)

Three blocks \(m_{1}, m_{2}\) and \(m_{3}\) of masses \(8 \mathrm{~kg}, 3 \mathrm{~kg}\) and \(1 \mathrm{~kg}\) are placed in contact on a smooth surface. Forces \(F_{1}=140 \mathrm{~N}\) and \(F_{2}=20 \mathrm{~N}\) are acting on blocks \(m_{1}\) and \(m_{3}\), respectively, as shown. The reaction between blocks \(m_{2}\) and \(m_{3}\) is (A) \(2.5 \mathrm{~N}\) (B) \(7.5 \mathrm{~N}\) (C) \(22.5 \mathrm{~N}\) (D) \(30 \mathrm{~N}\)

A block \(A\) of mass \(m\) is placed over a plank \(B\) of mass \(2 \mathrm{~m}\). Plank \(B\) is placed over a smooth horizontal surface. The co-efficient of friction between \(A\) and \(B\) is \(\frac{1}{2} .\) Block \(A\) is given a velocity \(v_{0}\) towards right. Acceleration of \(B\) relative to \(A\) is (A) \(\frac{g}{2}\) (B) \(g\) (C) \(\frac{3 g}{4}\) (D) Zero

A block of mass \(m\) is placed on a surface with a vertical cross-section given by \(y=\frac{x^{3}}{6} .\) If the co-efficient of friction is \(0.5\), the maximum height above the ground at which the block can be placed without slipping is: (A) \(\frac{1}{6} m\) (B) \(\frac{2}{3} m\) (C) \(\frac{1}{3} m\) (D) \(\frac{1}{2} m\)

A block of mass \(10 \mathrm{~kg}\) is placed on a rough inclined plane of inclination \(37^{\circ}\left(\tan 37^{\circ}=3 / 4\right)\). The co-efficient of friction between block and surface is \(0.4\). A horizontal force \(F=50 \mathrm{~N}\) is applied on the block, then \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\) (A) Acceleration of block is zero. (B) Acceleration of block is \(2.4 \mathrm{~m} / \mathrm{s}^{2}\) along the inclined plane. (C) Frictional force between block and surface is \(44 \mathrm{~N}\). (D) Frictional force between block and surface is \(20 \mathrm{~N}\).
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