A grapefruit falls from a tree and hits the ground 0.80 s later.
1 answer:
This question can be solved by using equations of motion.
A. The grapefruit dropped by "3.1 m".
B. The speed of the grapefruit, when it hit the ground was "7.8 m/s".
A.
We will use the second <em>equation of motion</em> here to find out the distance dropped by the grapefruit:
where,
h = height dropped = ?
vi = initial speed = 0 m/s
t = time period = 0.8 s
g = acceleration due to gravity = 9.81 m/s²
Therefore,
<u>h = 3.1 m</u>
<u />
B.
Now, we will use the first <em>equation of motion </em>to find out the final speed:
vf = 0 m/s + (9.81 m/s²)(0.8 s)
<u>vf = 7.8 m/s</u>
<u />
Learn more about <em>equations of motion</em> here:
brainly.com/question/20594939?referrer=searchResults
The attached picture shows the equations of motion in the horizontal and vertical directions.
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Answer:
The force of friction acting on block B is approximately 26.7N. Note: this result does not match any value from your multiple choice list. Please see comment at the end of this answer.
Explanation:
The acting force F=75N pushes block A into acceleration to the left. Through a kinetic friction force, block B also accelerates to the left, however, the maximum of the friction force (which is unknown) makes block B accelerate by 0.5 m/s^2 slower than the block A, hence appearing it to accelerate with 0.5 m/s^2 to the right relative to the block A.
To solve this problem, start with setting up the net force equations for both block A and B:
where forces acting to the left are positive and those acting to the right are negative. The friction force F_fr in the first equation is due to A acting on B and in the second equation due to B acting on A. They are opposite in direction but have the same magnitude (Newton's third law). We also know that B accelerates 0.5 slower than A:
Now we can solve the system of 3 equations for a_A, a_B and finally for F_fr:
The force of friction acting on block B is approximately 26.7N.
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