Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 431

A spring gun of spring constant \(90 \times 10^{2} \mathrm{~N} / \mathrm{M}\) is compressed \(4 \mathrm{~cm}\) by a ball of mass \(16 \mathrm{~g}\). If the trigger is pulled, calculate the velocity of the ball. (A) \(60 \mathrm{~m} / \mathrm{s}\) (B) \(3 \mathrm{~m} / \mathrm{s}\) (C) \(90 \mathrm{~m} / \mathrm{s}\) (D) \(30 \mathrm{~m} / \mathrm{s}\)

Short Answer

Expert verified
The correct answer is (D) \(30 \mathrm{~m} / \mathrm{s}\).

Step by step solution

01

Identify the given values and the unknown

In this problem, we are given: - Spring constant (k) = 90 × 10² N/m - Compression distance (x) = 4 cm (convert to meters) - Mass of the ball (m) = 16 g (convert to kg) - Initial velocity of the ball (vi) = 0 (since the ball is at rest) The unknown we need to find is the final velocity of the ball (vf) after the trigger is pulled.
02

Convert the given values to their appropriate SI units

First, we need to convert the compression distance (x) and the mass of the ball (m) to their appropriate SI units (meters and kilograms, respectively): - x = 4 cm = 0.04 m - m = 16 g = 0.016 kg
03

Find the potential energy stored in the compressed spring

Now, let's find the potential energy (PE) stored in the compressed spring using the formula: PE = 1/2 × k × x² PE = 1/2 × (90 × 10² N/m) × (0.04 m)² PE = 72 J (Joules)
04

Apply the conservation of energy principle

As per the conservation of energy principle, the potential energy stored in the compressed spring will convert to the kinetic energy (KE) of the ball when it is released. Thus, we can write: PE = KE Since the formula for kinetic energy is KE = 1/2 × m × v², we can equate the potential and kinetic energies: 72 J = 1/2 × 0.016 kg × vf²
05

Solve for the final velocity of the ball (vf)

Now, let's solve for the final velocity of the ball (vf): 72 J = 1/2 × 0.016 kg × vf² vf² = (72 × 2) / 0.016 vf² = 9000 vf = \( \sqrt{9000} \) vf ≈ 30 m/s The velocity of the ball after the trigger is pulled is approximately 30 m/s. Therefore, the correct answer is (D) \(30 \mathrm{~m} / \mathrm{s}\).

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

How much is the work done in pulling up a block of wood weighing $2 \mathrm{KN}\( for a length of \)10 \mathrm{~m}$ on a smooth plane inclined at an angle of \(30^{\circ}\) with the horizontal? (A) \(1.732 \mathrm{KJ}\) (B) \(17.32 \mathrm{KJ}\) (C) \(10 \mathrm{KJ}\) (D) \(100 \mathrm{KJ}\)

A force of \((2 \mathrm{i} \wedge+3 j \wedge-\mathrm{k} \wedge) \mathrm{N}\) acts on a body for 5 second, produces a displacement of $(3 i \wedge+5 j \wedge+k \wedge)$. What was the power used? (A) \(4 \mathrm{~W}\) (B) \(20 \mathrm{~W}\) (C) \(21 \mathrm{~W}\) (D) \(4.2 \mathrm{~W}\)

A uniform chain of length \(2 \mathrm{~m}\) is kept on a table such that a length of \(50 \mathrm{~cm}\) hangs freely from the edge of the table. The total mass of the chain is \(5 \mathrm{~kg}\). What is the work done in pulling the entire chain on the table. $\left(\mathrm{g}=10 \mathrm{~m} \mathrm{~s}^{2}\right)$ (A) \(7.2 \mathrm{~J}\) (B) \(3 \mathrm{~J}\) (C) \(4.6 \mathrm{~J}\) (D) \(120 \mathrm{~J}\)

A mass of \(\mathrm{M} \mathrm{kg}\) is suspended by a weight-less string, the horizontal force that is required to displace it until the string makes an angle of \(60^{\circ}\) with the initial vertical direction is (A) \(\mathrm{Mg} / \sqrt{3}\) (B) \(\mathrm{Mg} \cdot \sqrt{2}\) (C) \(\mathrm{Mg} / \sqrt{2}\) (D) \(\mathrm{Mg} \cdot \sqrt{3}\)

A ball dropped from a height of \(4 \mathrm{~m}\) rebounds to a height of $2.4 \mathrm{~m}$ after hitting the ground. Then the percentage of energy lost is (A) 40 (B) 50 (C) 30 (D) 600
See all solutions

Recommended explanations on English Textbooks

Language and Social Groups

Read Explanation

Prosody

Read Explanation

Rhetoric

Read Explanation

Multiple Choice Questions

Read Explanation

Text Comparison

Read Explanation

Synthesis Essay

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.

Sign-up for free