The work done on the box by the weight of the box is 1.8 Joule
<h3>Further explanation</h3>
<em>Let's recall </em><em>Kinetic Energy</em><em> and </em><em>Work </em><em>Formula as follows:</em>
<em>where:</em>
<em>Ek = Kinetic Energy ( Joule )</em>
<em>W = Work ( Joule )</em>
<em>m = mass of the object ( kg )</em>
<em>v = speed of the object ( m/s )</em>
<em>F = magnitude of force ( N )</em>
<em>d = displacement ( m )</em>
Let us now tackle the problem !
<u>Given:</u>
weight of the box = w = 2.0 N
angle of inclined plane = θ = 30°
normal force = N = 1.7 N
coefficient of kinetic friction = μ = 0.30
displacement of the box = d = 1.8 m
<u>Asked:</u>
work done by weight = W_w = ?
work done by normal force = W_n = ?
work done by the force of kinetic friction = W_fk = ?
<u>Solution:</u>
<em>We could calculate work done by weight as follows:</em>
<em>Because the direction of displacement is perpendicular to the direction of normal force , then we could calculate the work done by normal as follows:</em>
<em>Finally, we could calculate the work done by friction as follows:</em>
<h3>Learn more</h3>
<h3>Answer details</h3>
Grade: High School
Subject: Physics
Chapter: Dynamics