You are here

$W=8.1 \times 10^3J$

Giancoli 6th Edition, Chapter 6, Problem 4

(2:30)

Chapter 6, Problem 4 is solved.

View sample solution

Transcript for this Giancoli solution

Let's draw a picture of this crate to see the forces involved. Here's the ground. We put the crate on there. So the movers are pushing the crate at constant speed, and there's friction involved at this question. So we have gravity straight down, equal to 'mg' opposite that is normal force of equal size, because it's not moving vertically and so the normal force equals the gravity force and its 'mg'. That's going to be important for calculating the friction. If this is the direction of the friction force, label that friction. And then we'll have the applied force by the movers. Now that's the same length as the friction force. Let's give that the same length, because it's not accelerating. There's no net force. This is the force applied by the movers. Okay, so we'll say that the friction force is mu times 'Fn' that's the formula for friction. And that is mu times 'mg' since the normal force is mass times gravity acceleration due to gravity. And then we'll say that the applied force must equal the friction force, because it's not accelerating horizontally. And so the applied force is also mu 'mg'. And the work done will be that applied force times the distance and that's mu 'mg' times the distance that they move the crate. And that's going to work out to 0.50 that's the coefficient of friction times 160 kilograms times 9.8 newtons per kilogram, times 10.3 meters that they move the crate. And this is # times 10 to the # joules of work done by the movers. In the 5th Edition, you have a coefficient of friction of 0.7 and the crate is 150 Kilograms, and they move it 12.3 Meters giving a 5th Edition answer of # times 10 to the # joules.