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
When we need to calculate the resultant value of them.......
Answer:
cola forms so slowly that we could use up the supply that exists now
We will apply the equation:2as = v² - u²v = √(2as + u²)v = √(2 x 4 x 400 + 13²)v = 58 m/s
