Posted by Amisha on Monday, April 9, 2012 at 12:15pm.
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