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A flight attendant pulls her 75.8 N flight bag
a distance of 256 m along a level airport floor
at a constant velocity. The force she exerts is
47.0 N at an angle of 62.4◦
above the horizontal.
a) Find the work she does on the flight bag.
Answer in units of J.

Respuesta :

Rodiak
Let's start by drawing a free body diagram. We have two forces acting on a bag: pulling force and a friction force.

Pulling force is acting under some angle with respect to a ground. This means that we can find horizontal and vertical component of a force. We are only interested in horizontal component as the bag is moving over a horizontal floor.

We use this formula to find the component:
[tex]cos \alpha = \frac{horizontal component}{pulling force} \\ horizontal component = cos \alpha *pulling force \\ horizontal component = 21.8N[/tex]

The First Newton law says "if the body is moving at a constant speed the net force acting on an object is zero". This means that horizontal component of a pulling force and a friction force are equal.


Formula for work is:
[tex]W=F*s[/tex]

Where F is force causing the body to move, and s is a distance traveled.
[tex]W=21.8N*256m \\ W=5580.8J[/tex]
This is our work.

Why did we ignore vertical component?
If we calculate it we will get the value F=41.7N. This is less than the weight of a bag and it is not enough to cause a vertical movement of a bag.
Ver imagen Rodiak
a) work done = force x distance x cos(theta) where theta is the angle between the force and direction of motion

work done = 31.3 cos 52.6 x 278 = 5285J

b) the bag travels at a constant velocity, which means there is no change in kinetic energy; the work energy theorem tells us that total work done on an object equals the change in kinetic energy, therefore the total work done on the bag is zero. this means that friction does work that is equal and opposite to the work done by the person; work done by friction =-5285J

c) knowing the bag moves at a constant velocity, we know the sum of forces is zero

in the horizontal direction, this is the horizontal component of the pulling force and friction, or

P cos(theta)=friction
friction = u mg; Pcos(theta)=31.3 cos52.6;

mg = 70.8N, so we have

u = Pcos(theta)/mg = 31.3cos52.6/70.8
u=0.27

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