Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 3: Problem 13

A block of mass \(5 \mathrm{~kg}\) hangs by a string from the ceiling. cant copy image. The force the ceiling exerts on the string is most nearly: (A) \(1 \mathrm{~N}\), acting up (B) \(5 \mathrm{~N}\), acting down (C) \(50 \mathrm{~N}\), acting down (D) \(50 \mathrm{~N}\), acting up (E) \(100 \mathrm{~N}\), acting up

Short Answer

Expert verified
Answer: (D) \(50 \mathrm{~N}\), acting up (Note: The exact answer is 49 N, but we choose the closest provided option).

Step by step solution

01

Identify the forces acting on the block

First, we need to identify the forces acting on the 5 kg block. There are two forces acting on it: the force due to gravity (weight) and the tension force due to the string.
02

Calculate the force due to gravity (weight) on the block

To calculate the gravitational force or the weight of the block, we can use the formula: Weight = Mass × Acceleration due to gravity Here, the mass of the block is 5 kg and the acceleration due to gravity is approximately 9.8 m/s². So, the calculation becomes: Weight = 5 kg × 9.8 m/s² = 49 N (rounded to one decimal place) The weight of the block acts downward. [Note: Remember that the weight (force due to gravity) is not the same as the mass.]
03

Determine net force acting on the block

Since the block is in equilibrium (it is not accelerating), the net force acting on the block must be zero. According to Newton's second law of motion, net force can be calculated as: Net force = Mass × Acceleration Using the known mass of the block (5 kg) and the fact that it is not accelerating (acceleration = 0), we get: Net force = 5 kg × 0 = 0 N
04

Apply Newton's Third Law to find the force exerted by the ceiling on the string

According to Newton's Third Law, for every action, there is an equal and opposite reaction. In the context of this problem, the action is the tension force exerted by the string on the block, and the reaction is the force exerted by the ceiling on the string. Since the net force acting on the block is 0, the force exerted by the ceiling on the string must be equal and opposite to the weight of the block. In this case, the force exerted by the ceiling on the string is 49 N, acting upwards. The closest answer choice would be: (D) \(50 \mathrm{~N}\), acting up It is important to note that the exact answer is 49 N and not 50 N, but since the provided options do not have 49 N, we choose the closest value.

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

A crate of mass \(m\) is released from rest at the top of a stationary ramp of height \(h\), whose length along the curve is \(s\), as shown. Assuming that the ramp is frictionless, in calculating the speed of the crate \(v\) at the bottom of the ramp, one can assume: (A) \(h=v t\) (B) \(m g h=1 / 2 m v^2\) (C) \(s=1 / 2 a t^2\) (D) \(v^2=2 a s\) (E) \(h=1 / 2 g t^2\)

A \(60 \mathrm{~kg}\) passenger is traveling in a car at \(30 \mathrm{~m} / \mathrm{s}\) when suddenly a tree steps onto the road. The car crashes and the air bags deploy. If the passenger must be stopped within 1 meter, the force on the passenger is nearly (A) \(270 \mathrm{~N}\) (B) \(1800 \mathrm{~N}\) (C) \(2700 \mathrm{~N}\) (D) \(18,000 \mathrm{~N}\) (E) \(27,000 \mathrm{~N}\)

A spring obeys Hooke's law. If an amount of work \(W\) is required to stretch the spring a length \(x\) beyond its unstretched length, how much work does it take to stretch it to \(3 x\) ? (A) \(W\) (B) \(3 W\) (C) \(6 W\) (D) \(9 \mathrm{~W}\) (E) \(27 W\)

You find yourself stranded on planet Alpha again, which you remember is half as dense as Earth but which has a radius three times that of Earth's. What is the escape velocity \(v\) of Alpha compared to Earth's escape velocity? (A) \(2 / 3\) (B) The same (C) \(3 / 2\) (D) \(3 / \sqrt{2}\) (E) 6

cant copy graph Two students, Alice and Bob, decide to compute the power that the Earth's gravitational field expends on a block of mass \(m\) as the block slides down a frictionless inclined plane. Alice reasons: "The gravitational force pulling the block down the incline is \(F=m g \sin \theta\). The block's velocity at any given height \(h\) from the top of the incline is \(v=\sqrt{2 g h}\). Power is defined as force \(\times\) velocity. Therefore, the power is \(P=m g \sin \theta \sqrt{2 g h}\)." Bob reasons: "Power is \(\Delta W / \Delta t\). By the work-energy theorem, the change in work is the change in kinetic energy, but without friction \(\Delta W=m g h\). The change in time is \(\Delta t=\Delta v / a, \Delta v=\sqrt{2 g h}\) and \(a=g \sin \theta\). So \(\Delta t=\sqrt{2 g h} /(g \sin \theta)\). Therefore, \(P=m g h \times \frac{g \sin \theta}{\sqrt{2 g h}}=m g \sin \theta \sqrt{\frac{g h}{2}}\)." Alice and Bob look at each other and scratch their heads. Who is correct? \(\begin{array}{llll}\text { i. Alice } & \text { ii. Bob } & \text { iii. Neither iv. Both } & \text { v. The problem is imprecisely worded. }\end{array}\) (A) \(\mathrm{i}\) (B) ii (C) iii (D) iv (E) iv and \(v\)
See all solutions

Recommended explanations on Physics Textbooks

Measurements

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Classical Mechanics

Read Explanation

Fields in Physics

Read Explanation

Solid State Physics

Read Explanation

Magnetism and Electromagnetic Induction

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