Answer
given,
mass of the package = 12 kg
slides down distance = 2 m
angle of inclination = 53.0°
coefficient of kinetic friction = 0.4
a) work done on the package by friction is
W_f = -μk R d
= -μk (mg cos 53°)(2.0)
=-(0.4)(8.0 )(9.8)(cos 53°)(2.0)
= -37.75 J
b)
work done on the package by gravity is
W_g = m (g sin 53°) d
= (8.0 )(9.8 )(sin 53°)(2.0 )
=125.23 J
c)
the work done on the package by the normal force is
W_n = 0
d)
the net work done on the package is
W = -37.75 + 125.23 + 0
W = 87.84 J
The work done on the package by friction is equal to -37.75 Joules.
Given the following data:
Mass of package = 12.0 kg
Distance = 2.00 m
Angle of inclination = 53.0°
Coefficient of kinetic friction = 0.4.
Mathematically, the work done on the package by friction is given by this formula:
[tex]W_f = -\mu_k R d\\\\W_f = -\mu_k (mg cos\theta) d \\\\W_f =-0.40 \times (12.0 \times 9.8 \times cos53.0)\times 2.00\\\\W_f = -37.75\;Joules[/tex]
Mathematically, the work done on the package by gravity is given by this formula:
[tex]W_g = m g sin \theta d\\\\W_g = 12.0 \times 9.8 \times sin 53 \times 2.00 \\\\W_g = 125.23\;Joules[/tex]
In this scenario, the work done on the package by the normal force is equal to zero (0) Joules because the force is perpendicular.
[tex]W_{net}=W_f +W_g +W_N\\\\W_{net}=-37.75 + 125.23 + 0\\\\W_{net}=87.84 \;Joules[/tex]
Read more on work done here: brainly.com/question/22599382
