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MAXImum [283]
3 years ago
8

A classmate is attempting to draw a free body diagram for a box being pushed across the floor at a constant speed. The image bel

ow is what
they have drawn so far.
Which (if any) of their forces have been drawn incorrectly
O A. The applied force should be the same size as the friction force
B. There is nothing wrong with the free body diagram
OC. The normal force should be larger than the force and gravity
OD. The friction force should be in the same direction as the applied force

Physics
1 answer:
aivan3 [116]3 years ago
7 0

Answer:

A. The applied force should be the same size as the friction force

Explanation:

Whenever we apply a force to an object it moves if the force applied to that object is unbalanced and there is no force or a lesser force to counter it. According to Newton's Second Law of motion, when an unbalanced force is applied to an object it produces an acceleration in the object in its own direction. So, the two forces acting on this box are the frictional force and the applied force in horizontal direction. In order to move the box at constant speed, the applied force must first, overcome the frictional force, so the object can start its motion. Since, the motion has constant velocity, it means no acceleration. So, the force must be balanced in order to avoid acceleration as a consequence of Newton's Second Law of motion. Therefore, the correction in this case will be:

<u>A. The applied force should be the same size as the friction force</u>

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<h2>Answer: x=125m, y=48.308m</h2>

Explanation:

This situation is a good example of the projectile motion or parabolic motion, in which we have two components: x-component and y-component. Being their main equations to find the position as follows:

x-component:

x=V_{o}cos\theta t   (1)

Where:

V_{o}=54.5m/s is the projectile's initial speed

\theta=35\° is the angle

t=2.80s is the time since the projectile is launched until it strikes the target

x  is the final horizontal position of the projectile (the value we want to find)

y-component:

y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}   (2)

Where:

y_{o}=0  is the initial height of the projectile (we are told it  was launched at ground level)

y  is the final height of the projectile (the value we want to find)

g=9.8m/s^{2}  is the acceleration due gravity

Having this clear, let's begin with x (1):

x=(54.5m/s)cos(35\°)(2.8s)   (3)

x=125m   (4)  This is the horizontal final position of the projectile

For y (2):

y=0+(54.5m/s)sin(35\°)(2.8s)-\frac{(9.8m/s^{2})(2.8s)^{2}}{2}   (5)

y=48.308m   (6)  This is the vertical final position of the projectile

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