A 12 n cart is moving on a horizontal surface with a coefficient of kinetic friction of 0.20. what force of friction must be ove
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Answer:
68 m
Explanation:
Given that the horizontal velocity of the birdman = 17 m/s and
the height, h= 78 m.
The gravitational force is acting in the downward direction, so it will not change the horizontal speed.
The horizontal speed will remains be constant and will be equal to the initial horizontal speed of the turd.
Initially, the turd was also flying horizontally with the birdman, so the initial velocity of the turd is the same as the horizontal velocity of the birdman, i.e In the horizontal direction, m/s.
In the vertical direction, u = 0,
The distance to be traveled, in the direction of application of force, is equals to the height of the turd, i.e
s= 78 m
Let t be the time taken to cover a distance of 78 m.
Now, applying the equation of motion in the vertical direction,
where u is the initial velocity and a is the acceleration due to gravity in the direction of displacement,s.
Here, , so
seconds.
Hence, the time taken to reach the ground is 4 seconds.
As the horizontal speed, m/s, is constant throughout the journey, so
the horizontal distance covered by turd
m.
So, the distance of landing from the start of the field is 68 m as the birdman releases a turd directly above the start of the field.
Hence, the robot must hold the bucket at a distance of 68 m from the start of the field.