/*! 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} Q42P A worker pushed a 27 kg block ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A worker pushed a 27 kgblock9.2 malong a level floor at constant speed with a force directed32°below the horizontal. If the coefficient of kinetic friction between block and floor was 0.20, what were (a) the work done by the worker’s force and (b) the increase in thermal energy of the block– floor system?

Short Answer

Expert verified
  1. The work done by the force of worker is W=5.6×102J
  2. The increase in thermal energy of the block–floor system is ∆Eth=5.6×102J

Step by step solution

01

Step 1: Given Data

The mass of block is,m=27kg.

The displacement of a block is d=9.2m.

The applied force is directed32°below the horizontal.

The coefficient of kinetic friction is, μ=0.20.

02

Determining the concept

Use the equation of the work done related with force and displacement. Calculate the force applying Newton’s second law of motion. The thermal energy gets created due to the friction force, so the increase in thermal energy is equal to the work done against the friction force.

Formulae are as follow:

W=FdcosθFnet=ma

where, m is mass, a is an acceleration, d is displacement, F is force and W is work done.

03

Step 3(a): Determining the work done by the force of the worker

Draw a free body diagram for the block,

From this figure, applying Newton’s second law of motion to the horizontal direction,

Fnet=maFcos32-f=ma

But,

a=0,

So,

Fcos32-μkN=0Fcos32=μkmg+Fsin32

Solving this equation for applied force F,

F=μkmgcos32-μksin32whereμ=0.2,m=27kgF=71.39N

The equation for work done is,

W=Fdcosθ

So,

W=71.39×9.2×cos32W=557J≈5.6×102J

Hence, the work done by the force of worker is W=5.6×102J

04

Step 4(b): Determining the increase in thermal energy of the block–floor system

The change in thermal energy is given by,

∆Eth=fd∆Eth=μkmg+Fsin32×d∆Eth=0.2027×9.81+71.39sin32×9.2∆Eth=557J≈5.6×102J

Hence, the increase in thermal energy of the block–floor system is ∆Eth=5.6×102J

Therefore, the work done by the force of the worker and the increase in the thermal energy of the block-floor system can be found using the equation of work.

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!

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

In Fig. 8-55, a block slides along a path that is without friction until the block reaches the section of length L = 0.75 m, which begins at height h = 2.0 m on a ramp of angle θ=30° . In that section, the coefficient of kinetic friction is0.40. The block passes through point A with a speed of 8.0m/s. If the block can reach point B (where the friction ends), what is its speed there, and if it cannot, what is its greatest height above A?

A factory worker accidentally releases a 180 kgcrate that was being held at rest at the top of a ramp that is 3.7 m long and inclined at 39°to the horizontal. The coefficient of kinetic friction between the crate and the horizontal factory floor is 0.28. (a) How fast is the crate moving as it reaches the bottom of the ramp? (b) How far will it subsequently slide across the floor? (Assume that the crate’s kinetic energy does not change as it moves from the ramp onto the floor.) (c) Do the answers to (a) and (b) increase, decrease, or remain the same if we halve the mass of the crate?

Two blocks, of masses M = 2.0 kgand 2M, are connected to a spring of spring constantk = 200 N/mthat has one end fixed, as shown in Fig. 8-69. The horizontal surface and the pulley has negligible mass. The blocks are released from rest with the spring relaxed. (a) What is the combined kinetic energy of the two blocks when the hanging block has fallen 0.090m? (b) What is the kinetic energy of the hanging block when it has fallen that 0.090m ?(c) What maximum distance does the hanging block fall before momentarily stopping?

In Fig. 8-18, a horizontally moving block can take three frictionless routes, differing only in elevation, to reach the dashed finish line. Rank the routes according to (a) the speed of the block at the finish line and (b) the travel time of the block to the finish line, greatest first.

In Fig.8.57, a block is released from rest at height d =40 cmand slides down a frictionless ramp and onto a first plateau, which has lengthand where the coefficient of kinetic friction is 0.50. If the block is still moving, it then slides down a second frictionless ramp through height d/2and onto a lower plateau, which has length d/2and where the coefficient of kinetic friction is again0.50. If the block is still moving, it then slides up a frictionless ramp until it (momentarily) stops. Where does the block stop? If its final stop is on a plateau, state which one and give the distance Lfrom the left edge of that plateau. If the block reaches the ramp, give the height Habove the lower plateau where it momentarily stops.

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.