/*! 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 48 A block slides down an incline. ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A block slides down an incline. As it moves from point \(A\) to point \(B\), which are \(5.0 \mathrm{~m}\) apart, force \(\vec{F}\) acts on the block, with magnitude \(2.0 \mathrm{~N}\) and directed down the incline. The magnitude of the frictional force acting on the block is \(10 \mathrm{~N}\). If the kinetic energy of the block increases by \(35 \mathrm{~J}\) between \(A\) and \(B\), how much work is done on the block by the gravitational force as the block moves from \(A\) to \(B ?\)

Short Answer

Expert verified
The work done by the gravitational force is \(-25 \, J\).

Step by step solution

01

Identify Given Information

We are given the following details:- Distance between points A and B is \(5.0 \ m\).- Force \(\vec{F}\) acting on the block is \(2.0 \ N\) down the incline.- Frictional force magnitude is \(10 \ N\).- The increase in kinetic energy is \(35 \ J\).
02

Understand Work and Energy Relation

The work-energy principle states that the work done on an object is equal to the change in its kinetic energy. Given:\[\Delta KE = W_{total} = 35 \, J\]
03

Expressions for Work by Forces

The total work \(W_{total}\) done on the block is by the gravitational force, the applied force \(\vec{F}\), and the frictional force, expressed as:\[W_{total} = W_g + W_F - W_f\] where:- \(W_g\) is the work done by gravity.- \(W_F\) is the work done by force \(\vec{F}\).- \(W_f\) is the work done by the frictional force.
04

Calculate Work Done by Force \(\vec{F}\)

The work done by force \(\vec{F}\) is calculated as:\[W_F = F \cdot d \cdot \cos(\theta)\]Where \(\theta = 0^{\circ}\) since it's along the direction of motion:\[W_F = 2 \, N \times 5 \, m = 10 \, J\]
05

Calculate Work Done by Frictional Force

The work done by the frictional force is:\[W_f = F_{friction} \cdot d \cdot \cos(180^{\circ})\]Because friction opposes motion, \(\cos(180^{\circ}) = -1\):\[W_f = 10 \, N \times 5 \, m \times -1 = -50 \, J\]
06

Setup Equation for Gravitational Work

Substitute the known values into the work-energy equation to find \(W_g\):\[35 \, J = W_g + 10 \, J - (-50 \, J)\]Simplifying gives:\[35 \, J = W_g + 10 \, J + 50 \, J\]
07

Calculate Work Done by Gravitational Force

Rearrange the equation to solve for \(W_g\):\[35 \, J = W_g + 60 \, J\]\[W_g = 35 \, J - 60 \, J\]\[W_g = -25 \, J\]Thus, the work done by the gravitational force is \(-25 \, J\).

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 91Ó°ÊÓ!

Key Concepts

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

Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. It's a key concept in physics and is expressed by the formula: \[KE = \frac{1}{2}mv^2\] where:
  • \(m\) is the mass of the object
  • \(v\) is its velocity
This equation demonstrates that the kinetic energy depends both on mass and the square of the velocity, making velocity a significant factor. For our exercise, the block’s kinetic energy increases by 35 J as it travels from point A to B on the incline. This increase signifies that, overall, the block accelerated or picked up speed as it moved, which is due to the net work done on it. Understanding how forces like gravity and applied forces influence kinetic energy is crucial here. They can either increase or decrease it, depending on their direction relative to the object's motion.
Gravitational Force
Gravitational force is a natural phenomenon where objects with mass attract one another. On Earth, this is often experienced as the force that gives weight to objects, pulling them towards the center of the planet. In the context of our exercise, gravitational force acts on the block as it slides down the incline. This force contributes to the block’s movement and influences its kinetic energy change. The work done by gravity can be calculated by understanding its effect over a distance, along the inclined path, which is determined by: \[W_g = mgh\] Here, \(W_g\) is the work done by gravity, \(m\) is the mass of the block, \(g\) is the acceleration due to gravity, and \(h\) is the vertical displacement. In our scenario, the gravitational force does negative work amounting to \(-25 \, J\), meaning it acts somewhat against the direction of the net motion, possibly due to the high resistance from the frictional force and applied forces.
Frictional Force
Frictional force is a resistive force that opposes the motion of an object. It acts along surfaces and is often encountered when two surfaces are in contact. The direction of frictional force is always opposite to the direction of movement. In our case, as the block moves down the incline, the frictional force opposing it is 10 N. This force plays a significant role in affecting the block's overall movement. The work done by friction is calculated by this formula: \[W_f = F_{friction} \times d \times \cos(180^{\circ})\] Since this work acts oppositely to the direction of motion, it has a negative value, specifically -50 J in our problem. Understanding friction helps explain why, despite the forces acting down the incline, not all the kinetic energy expected from the forces' work appears as the increase in energy, as some energy is 'lost' due to friction. This 'loss' refers to the energy used to overcome this resistance as the block moves down.

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 \(2.0 \mathrm{~kg}\) breadbox on a frictionless incline of angle \(\theta=\) \(40^{\circ}\) is connected, by a cord that runs over a pulley, to a light spring of spring constant \(k=120 \mathrm{~N} / \mathrm{m}\), as shown in Fig. \(8-41 .\) The box is released from rest when the spring is unstretched. Assume that the pulley is massless and frictionless. (a) What is the speed of the box when it has moved \(10 \mathrm{~cm}\) down the incline? (b) How far down the incline from its point of release does the box slide before momentarily stopping, and what are the (c) magnitude and (d) direction (up or down the incline) of the box's acceleration at the instant the box momentarily stops?

A \(0.42 \mathrm{~kg}\) shuffleboard disk is initially at rest when a player uses a cue to increase its speed to \(4.2 \mathrm{~m} / \mathrm{s}\) at constant acceleration. The acceleration takes place over a \(2.0 \mathrm{~m}\) distance, at the end of which the cue loses contact with the disk. Then the disk slides an additional \(12 \mathrm{~m}\) before stopping. Assume that the shuffleboard court is level and that the force of friction on the disk is constant. What is the increase in the thermal energy of the disk-court system (a) for that additional \(12 \mathrm{~m}\) and \((\mathrm{b})\) for the entire \(14 \mathrm{~m}\) distance? (c) How much work is done on the disk by the cue?

A \(1.50 \mathrm{~kg}\) snowball is shot upward at an angle of \(34.0^{\circ}\) to the horizontal with an initial speed of \(20.0 \mathrm{~m} / \mathrm{s}\). (a) What is its initial kinetic energy? (b) By how much does the gravitational potential energy of the snowball-Earth system change as the snowball moves from the launch point to the point of maximum height? (c) What is that maximum height?

A \(68 \mathrm{~kg}\) sky diver falls at a constant terminal speed of \(59 \mathrm{~m} / \mathrm{s}\). (a) At what rate is the gravitational potential energy of the Earth-sky diver system being reduced? (b) At what rate is the system's mechanical energy being reduced?

A machine pulls a \(40 \mathrm{~kg}\) trunk \(2.0 \mathrm{~m}\) up a \(40^{\circ} \mathrm{ramp}\) at constant velocity, with the machine's force on the trunk directed parallel to the ramp. The coefficient of kinetic friction between the trunk and the ramp is \(0.40\). What are (a) the work done on the trunk by the machine's force and (b) the increase in thermal energy of the trunk and the ramp?

See all solutions

Recommended explanations on Physics Textbooks

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.