posted by Amisha .
An 8.00kg package in a mail-sorting room slides 2.00 m down a chute that is inclined at 53.0 degrees below the horizontal. The coefficient of kinetic friction between the package and the chute's surface is 0.400.
Calculate the work done on the package by the normal force.What is the net work done on the package?
The "normal force" is always perpendicular to the direction of motion, and does no work. Zero.
The normal force is M*g*cos53 = 47.2 N
The friction force is 0.40 times the normal force, or 18.9 N
After sliding down a vertical distance
H = 2.00 sin 53 = 1.597 m,
Work done by gravity = MgH = 125.2 J
Work done by friction
= -18.9 * 2.00 = -37.8 J
Net work done on the package
= 87.4 J
Normal force is perpendicular to the displacement of the package; therefore, its work is zero.
The net work is equal to the difference between of the work of gravity and the work of the friction force
W(g) =m•g•h = m•g•s•sinα = 8•9.8•2•sin53 =125.2 J
W(fr) = F(fr) •s =k•N•s=k•m•g•cos α•s = 0.4•8•9.8•0.6•2 =37.75 J.
W = W(g)+ W(fr) =125.2 - 37.75= 87.45 J
Work done on the package by the Normal force
Normal force N is perpendicular to the displacement
WN = mg cos53 (d cos90 )
WN = 0 J
Work done by friction force, Wf = Ff (d cosθ)
where F f is friction force.,
Ffr = 0.4kN = 0.40*mg cos 53 =
Wf = (0.4)(8.0)(9.81)cos53(2.0) cos 180
Wf = 37.7 Joule
Work done by the gravity
Wg = – mgH
= – (8.0)(9.81)(1.59 m)
= – 125.2 Joule
So total of the work, Wnett = WN + Wf + WG
= 0 J + 37.7 J – 125.2 J = – 87.5 J
because the work in the direction slide down or moving downward