/*! 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 1 A force of \(10 \mathrm{~N}\) ac... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 1

A force of \(10 \mathrm{~N}\) acts on a mass of \(2 \mathrm{~kg}\) for three seconds. If the initial velocity was \(50 \mathrm{~ms}^{-1}\) what is the final velocity?

Short Answer

Expert verified
The final velocity is 65 m/s.

Step by step solution

01

Identify the known values

The force applied is 10 N, the mass of the object is 2 kg, the time duration is 3 seconds, and the initial velocity is 50 m/s.
02

Use Newton's Second Law

According to Newton鈥檚 Second Law, the acceleration can be calculated using the formula \(a = \frac{F}{m}\). Substitute the given values: \(a = \frac{10 \mathrm{~N}}{2 \mathrm{~kg}} = 5 \mathrm{~ms}^{-2}\).
03

Calculate the change in velocity

The change in velocity (鈭唙) can be found using the formula \(\Delta v = a \times t\). Substitute the values for acceleration and time: \(\Delta v = 5 \mathrm{~ms}^{-2} \times 3 \mathrm{~s} = 15 \mathrm{~ms}^{-1}\).
04

Find the final velocity

The final velocity (v_f) is the initial velocity (v_i) plus the change in velocity (鈭唙). Substitute the known values into the formula: \(v_f = v_i + \Delta v\). So, \(v_f = 50 \mathrm{~ms}^{-1} + 15 \mathrm{~ms}^{-1} = 65 \mathrm{~ms}^{-1}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Force and Mass
Newton's Second Law of Motion states that the force applied to an object is directly proportional to the acceleration of that object and inversely proportional to its mass. This is described by the formula: \[ F = ma \] Here, \'F\' is the force applied, \'m\' is the mass of the object, and \'a\' is the acceleration. When you apply a force of 10 N to a mass of 2 kg, you can calculate the acceleration easily.
You just need to use the formula: \[ a = \frac{F}{m} \] Substituting the given values, we get: \[ a = \frac{10 \: \textrm{N}}{2 \: \textrm{kg}} = 5 \: \textrm{ms}^{-2} \] This means the object will accelerate at a rate of 5 meters per second squared when a force of 10 N is applied to a mass of 2 kg.
Acceleration Calculation
Acceleration is a change in velocity over time. Once we know the acceleration, we can use it to find out how the object鈥檚 velocity changes over a specified time period. In the given problem, we have an acceleration of 5 \textrm{ms}^{-2} acting for a time duration of 3 seconds.
To calculate the change in velocity, we use the formula: \[ \Delta v = a \times t \] Here, \'\Delta v\' is the change in velocity, \'a\' is the acceleration, and \'t\' is the time.
Substitute the known values: \[ \Delta v = 5 \: \textrm{ms}^{-2} \times 3 \: \textrm{s} = 15 \: \textrm{ms}^{-1} \] Therefore, the object鈥檚 velocity changes by 15 meters per second over the period of 3 seconds due to the applied force.
Change in Velocity
To find the final velocity of the object, we need to consider its initial velocity and the change in velocity due to the applied force. The final velocity \'v_f\' can be calculated by adding the initial velocity \'v_i\' to the change in velocity \'\Delta v\'. This can be expressed with the following formula: \[ v_f = v_i + \Delta v \] In the given problem, the initial velocity is 50 \textrm{ms}^{-1}, and the change in velocity is 15 \textrm{ms}^{-1}.
Substitute these values into the formula: \[ v_f = 50 \: \textrm{ms}^{-1} + 15 \: \textrm{ms}^{-1} = 65 \: \textrm{ms}^{-1} \] Therefore, the final velocity of the object is 65 meters per second after a force of 10 N is applied for 3 seconds.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A pump raises water from a depth of \(10 \mathrm{~m}\) and discharges it horizontally through a pipe of \(0.1 \mathrm{~m}\) diameter at a velocity of \(8 \mathrm{~ms}^{-1} .\) Calculate the work done by the pump in one second. If the water impinges directly with the same velocity on a vertical wall, find the force exerted by the water on the wall if it is assumed that none of the water bounces back (Take \(g\) as \(9.81 \mathrm{~ms}^{-2}, \pi\) as \(3.142\) and the mass of \(1 \mathrm{~m}^{3}\) of water as \(1000 \mathrm{~kg}\) ).

18) Two particles of masses \(m\) and \(3 m\) are connected by a light inelastic string of length \(2 l\) which passes over a small smooth fixed peg. The particles are held in contact with the peg and then allowed, at the same instant, to fall from rest under gravity, one on either side of the peg. Prove that: (i) the speed of each particle just after the string tightens is \(\sqrt{(g l / 2)}\), (ii) the sudden tightening of the string causes a loss of energy equal to \(3 \mathrm{mgl}\), (iii) the lighter particle reaches the peg again after a total time \(\sqrt{61 / g}\).

Three particles \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\), each of mass \(m\), lie at rest on a smooth horizontal table. Light inextensible strings connect \(\mathrm{A}\) to \(\mathrm{B}\) and \(\mathrm{B}\) to \(\mathrm{C}\). The strings are just taut with the angle \(A B C=135^{\circ}\), when a blow of impulse \(J\) is applied to \(C\) in a direction parallel to \(\overrightarrow{\mathrm{AB}}\). Prove that A begins to move with speed \((J / 7) m\) and find the impulsive tension in the string \(\mathrm{BC}\).

The barrel of a gun of mass \(M\) resting on a smooth horizontal plane is elevated at an angle \(\alpha\) to the horizontal. The gun fires a shell of mass \(m\) and recoils with horizontal velocity \(U .\) If the velocity of the shell on leaving the gun has horizontal and vertical components \(v\) and \(w\) respectively, prove that \(w=(v+U) \tan \alpha\), and hence or otherwise prove that the initial inclination of the path of the shell to the horizontal is arctan \(\left[\left(1+\frac{m}{M}\right) \tan \alpha\right]\). Prove that the kinetic energy generated by the explosion is $$ \frac{U^{2}}{2 m}(M+m)\left(M \sec ^{2} \alpha+m \tan ^{2} \alpha\right) $$

A golf ball, initially at rest, is dropped on to a horizontal surface and bounces directly up again with velocity \(v\). If the coefficient of restitution between the ball and the surface is \(e\), show that the ball will go on bouncing for a time \(\frac{2 v}{g(1-e)}\) after the first impact. \(\left\\{\right.\) You may assume \(\left.1+e+e^{2}+e^{3}+\ldots .=(1-e)^{-1}\right\\}\). If the golf ball is dropped from a height of \(19.62 \mathrm{~m}\) and comes to rest 12 seconds later, find the value of \(e\). (Take \(g\) as \(9.81 \mathrm{~ms}^{-2}\) ).
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Pure Maths

Read Explanation

Probability and Statistics

Read Explanation

Applied Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Decision Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.