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
Answer:
(i) 208 cm from the pivot
(ii) Move further from the pivot
Explanation:
(i) Sum of the moments about the pivot of the seesaw is zero.
∑τ = Iα
(50 kg) (10 N/kg) (2.5 m) + (60 kg) (10 N/kg) x = 0
1250 Nm + 600 N x = 0
x = -2.08 m
Kenny should sit 208 cm on the other side of the pivot.
(ii) To increase the torque, Kenny should move away from the pivot.
The resistance would increase.
It is C) because that truck has more mass
