Friction

The work done is calculated as:

$W%20%3D%20F%20%5Ccdot%20d%20%5Ccdot%20%5Ccos%5Ctheta$ ,

where:

$F$ is the force applied,
$d$ is the distance moved, and
$%5Ctheta%20%3D%2045%5E%5Ccirc$ is the angle between the force and the direction of motion.

The force required to move the box at constant speed is equal to the weight of the box since friction is negligible:

$F%20%3D%20m%20%5Ccdot%20g%20%3D%2020%20%5Ctext%7B%20kg%7D%20%5Ccdot%209.8%20%5Ctext%7B%20m%2Fs%7D%5E2%20%3D%20196%20%5Ctext%7B%20N%7D.$

The work done is:

$W%20%3D%20196%20%5Ccdot%2020%20%5Ccdot%20%5Ccos%2045%5E%5Ccirc%20%3D%20196%20%5Ccdot%2020%20%5Ccdot%20%5Cfrac%7B1%7D%7B%5Csqrt%7B2%7D%7D%20%5Capprox%202790%20%5Ctext%7B%20J%7D.$

High-Yield Trend

Chapter Questions
1 MCQs

Official Solution

Friction

High-Yield Trend

Chapter Questions 1 MCQs

Official Solution

Chapter Questions
1 MCQs