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
Well, well, well, looks like we have a slippery situation here! Let's dive right into it, shall we?
To calculate the work done on the package by the normal force, we first need to find the normal force itself. In this case, the normal force is the force exerted by the surface perpendicular to the package. And since we have a sloping surface, we need to find the component of the normal force that acts in the direction of displacement.
Now, the angle of inclination is given as 53.0 degrees, so the component of the normal force acting in the direction of displacement is given by:
Normal Force * cos(theta),
where theta is the angle of inclination. So, the work done by the normal force is then:
Work = Normal Force * cos(theta) * 2.00 m.
Now let's tackle the net work done on the package. The net work done is the sum of all the forces on the package multiplied by the displacement. In this case, we have two forces at play: the force of gravity and the force of friction.
The work done by gravity is given by:
Work_gravity = m * g * h,
where m is the mass of the package, g is the acceleration due to gravity, and h is the vertical height the package slides down. In this case, the vertical height is given by:
h = 2.00 m * sin(theta),
where theta is the angle of inclination.
The work done by friction is given by:
Work_friction = -u * Normal Force * cos(theta) * 2.00 m,
where u is the coefficient of kinetic friction.
Finally, the net work done on the package is simply the sum of these two works:
Net Work = Work_gravity + Work_friction.
Now go ahead and plug in the numbers into these equations, calculate the values, and see what you get. Just remember to hold on tight and don't slide down any inclined chutes yourself while doing so!
To calculate the work done on the package by the normal force, we need to consider the forces acting on the package along the direction of the displacement.
The forces acting on the package are the gravitational force (mg), the normal force (N), and the force of kinetic friction (f). The work done by a force is given by the equation W = F * d * cos(theta), where F is the force, d is the displacement, and theta is the angle between the force and the displacement.
Since the normal force is perpendicular to the displacement, the angle between the normal force and the displacement is 90 degrees, and cos(theta) = cos(90) = 0. Therefore, the work done by the normal force is zero because cos(theta) = 0.
Now, let's calculate the net work done on the package. The net work done is equal to the work done by the gravitational force minus the work done by the force of kinetic friction.
1. Work done by the gravitational force:
W_gravity = mg * d * cos(theta)
Given: m = 8.00 kg, g = 9.8 m/s^2, d = 2.00 m
W_gravity = (8.00 kg * 9.8 m/s^2) * (2.00 m) * cos(0) = 156.8 J
2. Work done by the force of kinetic friction:
The force of kinetic friction is given by f = coefficient of kinetic friction * N, where N is the normal force.
The work done by the force of kinetic friction is given by W_friction = f * d * cos(theta), but since cos(theta) = 0, the work done by kinetic friction is also zero.
Therefore, the net work done on the package is equal to the work done by the gravitational force, which is 156.8 J.