A0.60-kg book slides on a horizontal table. the kinetic friction force on the book has magnitude 1.9 n . part a how much work is done on the book by friction during a displacement of 2.3 m to the left?

4.37 J

Explanation:

The work done by a constant force is given by:

where

F is the magnitude of the force

d is the distance through which the object has been moved

is the angle between the direction of the force and of the motion of the object.

In this problem:

- the magnitude of the force of friction is F = 1.9 N

- the distance is d = 2.3 m

- the angle is , because the direction of the friction is opposite to the direction of motion of the book

Therefore, the work done is

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west