The work done on the box by the weight of the box is 1.8 Joule
The work done on the box by the normal force is 0 Joule
The work done on the box by the force of kinetic friction is -0.918 Joule
<h3>Further explanation</h3>
<em>Let's recall </em><em>Kinetic Energy</em><em> Formula as follows:</em>
Ek = Kinetic Energy ( Joule )
m = mass of the object ( kg )
v = speed of the object ( m/s )
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_f = ?
<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:</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
Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant