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A constant force of frictionof 50 N is acting on body of mass 200kg moving initially with a speed of 15m/s? How does the body take to stop? What distance it will cover before coming to rest?
Verified Answer
A constant force of frictionof 50 N is acting on body of mass 200kg mo...
F=-50N
M=200kg
U=15m/s
F=Ma
a=F/M=-50/200=-0.25
Now using
V^2-U^2=2aS
0-(15)^2=2(-0.25)S
S=-225/-0.5=450m
As V=U+at
0=15-0.25t
t=-15/-0.25=60s
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Most Upvoted Answer
A constant force of frictionof 50 N is acting on body of mass 200kg mo...
Analysis:
To solve this problem, we need to use the concept of Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration.
Given:
- Mass of the body (m) = 200 kg
- Initial speed of the body (u) = 15 m/s
- Force of friction (F) = 50 N
Time taken to stop:
Using Newton's second law, we can find the acceleration (a) of the body:
F = m * a
50 = 200 * a
a = 50/200
a = 0.25 m/s^2
We know that acceleration is the rate of change of velocity, so we can use the following equation to find the time taken to stop (t):
v = u + a * t
0 = 15 + (-0.25) * t
-15 = -0.25t
t = 15/0.25
t = 60 seconds
Therefore, the body will take 60 seconds to stop.
Distance covered before coming to rest:
To find the distance covered (s) before coming to rest, we can use the equation:
s = ut + (1/2) * a * t^2
Substituting the given values:
s = 15 * 60 + (1/2) * 0.25 * (60)^2
s = 900 + (0.5) * 0.25 * 3600
s = 900 + 450
s = 1350 meters
Therefore, the body will cover a distance of 1350 meters before coming to rest.
Conclusion:
- The body will take 60 seconds to stop.
- It will cover a distance of 1350 meters before coming to rest.
To solve this problem, we need to use the concept of Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration.
Given:
- Mass of the body (m) = 200 kg
- Initial speed of the body (u) = 15 m/s
- Force of friction (F) = 50 N
Time taken to stop:
Using Newton's second law, we can find the acceleration (a) of the body:
F = m * a
50 = 200 * a
a = 50/200
a = 0.25 m/s^2
We know that acceleration is the rate of change of velocity, so we can use the following equation to find the time taken to stop (t):
v = u + a * t
0 = 15 + (-0.25) * t
-15 = -0.25t
t = 15/0.25
t = 60 seconds
Therefore, the body will take 60 seconds to stop.
Distance covered before coming to rest:
To find the distance covered (s) before coming to rest, we can use the equation:
s = ut + (1/2) * a * t^2
Substituting the given values:
s = 15 * 60 + (1/2) * 0.25 * (60)^2
s = 900 + (0.5) * 0.25 * 3600
s = 900 + 450
s = 1350 meters
Therefore, the body will cover a distance of 1350 meters before coming to rest.
Conclusion:
- The body will take 60 seconds to stop.
- It will cover a distance of 1350 meters before coming to rest.
Community Answer
A constant force of frictionof 50 N is acting on body of mass 200kg mo...
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A constant force of frictionof 50 N is acting on body of mass 200kg moving initially with a speed of 15m/s? How does the body take to stop? What distance it will cover before coming to rest?
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A constant force of frictionof 50 N is acting on body of mass 200kg moving initially with a speed of 15m/s? How does the body take to stop? What distance it will cover before coming to rest? for Class 9 2024 is part of Class 9 preparation. The Question and answers have been prepared according to the Class 9 exam syllabus. Information about A constant force of frictionof 50 N is acting on body of mass 200kg moving initially with a speed of 15m/s? How does the body take to stop? What distance it will cover before coming to rest? covers all topics & solutions for Class 9 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A constant force of frictionof 50 N is acting on body of mass 200kg moving initially with a speed of 15m/s? How does the body take to stop? What distance it will cover before coming to rest?.
A constant force of frictionof 50 N is acting on body of mass 200kg moving initially with a speed of 15m/s? How does the body take to stop? What distance it will cover before coming to rest? for Class 9 2024 is part of Class 9 preparation. The Question and answers have been prepared according to the Class 9 exam syllabus. Information about A constant force of frictionof 50 N is acting on body of mass 200kg moving initially with a speed of 15m/s? How does the body take to stop? What distance it will cover before coming to rest? covers all topics & solutions for Class 9 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A constant force of frictionof 50 N is acting on body of mass 200kg moving initially with a speed of 15m/s? How does the body take to stop? What distance it will cover before coming to rest?.
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