Q. Potential energy curve of a body of mass m = 2kg is shown in the figure, it is released from rest at point A. What is the maximum distance covered by it on the positive x-axis till it momentarily comes to rest from the origin?
#ANSWER THE QUESTION WITH PROPER EXPLANATION
#No Spamming Plz
Answers
The slope of gravitational potential energy and distance give,the maginitude of force.....
This can be easily understand by seeing the slope of that line(in figure),
it's slope=U/x=mgh/h=mg
which is the magnitude of force...
Now, let's move on question
In this,the coordinate of the straight line are
(2,32) and (0,0)
It's slope=(32-0)/2-0=16
From the above statement,we can say that,this will be the force....
16=m×a
a=16/2=8m/sec^2
when,the object will reach at the bottom then it's potential energy is totally converted into kinetic energy
mgh=(1/2)×m×v^2
16×2/2=v^2
v=4
.now,
initial velocity=4..(when it reach to the ground)
final velocity=0..(it have to come to rest)
acceleration=8
so, using third law of uniformly accelerated motion,we get
0^2=16-2×8×s
s=1m
hence the body will cover 1m distance before it will come to rest
Answer:
Explanation:
1 m is the answer.
