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A 50 - kg box rests on horizontal floor for which coefficient of friction is 0.3. If the box is subjected to a horizontal towing force of 400 N, what is its velocity after 5 sec from rest? (Take g = 10 m/s2)
- a)40 m/s
- b)30 m/s
- c)27.5 m/s
- d)25 m/s
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
A 50 - kg box rests on horizontal floor for which coefficient of frict...
Initially checking whether towing force can overcome the static friction force or not.
Friction force (f) = μN = μmg = 0.3 × 50 × 10 = 150 N
∴ Net force on box = 400 – 150 = 250 N
⇒ 250 = ma
Velocity after 5 second
V = u + at = 0 + 5 × 5 = 25 m/s
Most Upvoted Answer
A 50 - kg box rests on horizontal floor for which coefficient of frict...
Given data:
Mass of the box, m = 50 kg
Coefficient of friction, μ = 0.3
Towing force, F = 400 N
Acceleration due to gravity, g = 10 m/s^2
Time, t = 5 s
To find:
Velocity of the box after 5 seconds from rest
Firstly, we need to calculate the frictional force acting on the box. The frictional force can be found using the equation:
Frictional force, F_friction = μ × Normal force
The normal force acting on the box is equal to the weight of the box, which can be calculated as:
Normal force, N = mg
Where,
m = mass of the box
g = acceleration due to gravity
Substituting the values, we get:
Normal force, N = 50 kg × 10 m/s^2 = 500 N
Now, we can calculate the frictional force:
F_friction = 0.3 × 500 N = 150 N
The net force acting on the box can be found by subtracting the frictional force from the applied force:
Net force, F_net = F - F_friction
Substituting the values, we get:
F_net = 400 N - 150 N = 250 N
Next, we can calculate the acceleration of the box using Newton's second law of motion:
F_net = ma
Substituting the values, we get:
250 N = 50 kg × a
Simplifying, we get:
a = 250 N / 50 kg = 5 m/s^2
Finally, we can calculate the velocity of the box after 5 seconds using the equation of motion:
v = u + at
Where,
v = final velocity
u = initial velocity (0 m/s)
a = acceleration
t = time
Substituting the values, we get:
v = 0 + 5 m/s^2 × 5 s = 25 m/s
Therefore, the velocity of the box after 5 seconds from rest is 25 m/s, which corresponds to option D.
Mass of the box, m = 50 kg
Coefficient of friction, μ = 0.3
Towing force, F = 400 N
Acceleration due to gravity, g = 10 m/s^2
Time, t = 5 s
To find:
Velocity of the box after 5 seconds from rest
Firstly, we need to calculate the frictional force acting on the box. The frictional force can be found using the equation:
Frictional force, F_friction = μ × Normal force
The normal force acting on the box is equal to the weight of the box, which can be calculated as:
Normal force, N = mg
Where,
m = mass of the box
g = acceleration due to gravity
Substituting the values, we get:
Normal force, N = 50 kg × 10 m/s^2 = 500 N
Now, we can calculate the frictional force:
F_friction = 0.3 × 500 N = 150 N
The net force acting on the box can be found by subtracting the frictional force from the applied force:
Net force, F_net = F - F_friction
Substituting the values, we get:
F_net = 400 N - 150 N = 250 N
Next, we can calculate the acceleration of the box using Newton's second law of motion:
F_net = ma
Substituting the values, we get:
250 N = 50 kg × a
Simplifying, we get:
a = 250 N / 50 kg = 5 m/s^2
Finally, we can calculate the velocity of the box after 5 seconds using the equation of motion:
v = u + at
Where,
v = final velocity
u = initial velocity (0 m/s)
a = acceleration
t = time
Substituting the values, we get:
v = 0 + 5 m/s^2 × 5 s = 25 m/s
Therefore, the velocity of the box after 5 seconds from rest is 25 m/s, which corresponds to option D.
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A 50 - kg box rests on horizontal floor for which coefficient of friction is 0.3. If the box is subjected to a horizontal towing force of 400 N, what is its velocity after 5 sec from rest? (Take g = 10 m/s2)a)40 m/sb)30 m/sc)27.5 m/sd)25 m/sCorrect answer is option 'D'. Can you explain this answer?
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A 50 - kg box rests on horizontal floor for which coefficient of friction is 0.3. If the box is subjected to a horizontal towing force of 400 N, what is its velocity after 5 sec from rest? (Take g = 10 m/s2)a)40 m/sb)30 m/sc)27.5 m/sd)25 m/sCorrect answer is option 'D'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A 50 - kg box rests on horizontal floor for which coefficient of friction is 0.3. If the box is subjected to a horizontal towing force of 400 N, what is its velocity after 5 sec from rest? (Take g = 10 m/s2)a)40 m/sb)30 m/sc)27.5 m/sd)25 m/sCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 50 - kg box rests on horizontal floor for which coefficient of friction is 0.3. If the box is subjected to a horizontal towing force of 400 N, what is its velocity after 5 sec from rest? (Take g = 10 m/s2)a)40 m/sb)30 m/sc)27.5 m/sd)25 m/sCorrect answer is option 'D'. Can you explain this answer?.
A 50 - kg box rests on horizontal floor for which coefficient of friction is 0.3. If the box is subjected to a horizontal towing force of 400 N, what is its velocity after 5 sec from rest? (Take g = 10 m/s2)a)40 m/sb)30 m/sc)27.5 m/sd)25 m/sCorrect answer is option 'D'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A 50 - kg box rests on horizontal floor for which coefficient of friction is 0.3. If the box is subjected to a horizontal towing force of 400 N, what is its velocity after 5 sec from rest? (Take g = 10 m/s2)a)40 m/sb)30 m/sc)27.5 m/sd)25 m/sCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 50 - kg box rests on horizontal floor for which coefficient of friction is 0.3. If the box is subjected to a horizontal towing force of 400 N, what is its velocity after 5 sec from rest? (Take g = 10 m/s2)a)40 m/sb)30 m/sc)27.5 m/sd)25 m/sCorrect answer is option 'D'. Can you explain this answer?.
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A 50 - kg box rests on horizontal floor for which coefficient of friction is 0.3. If the box is subjected to a horizontal towing force of 400 N, what is its velocity after 5 sec from rest? (Take g = 10 m/s2)a)40 m/sb)30 m/sc)27.5 m/sd)25 m/sCorrect answer is option 'D'. Can you explain this answer?, a detailed solution for A 50 - kg box rests on horizontal floor for which coefficient of friction is 0.3. If the box is subjected to a horizontal towing force of 400 N, what is its velocity after 5 sec from rest? (Take g = 10 m/s2)a)40 m/sb)30 m/sc)27.5 m/sd)25 m/sCorrect answer is option 'D'. Can you explain this answer? has been provided alongside types of A 50 - kg box rests on horizontal floor for which coefficient of friction is 0.3. If the box is subjected to a horizontal towing force of 400 N, what is its velocity after 5 sec from rest? (Take g = 10 m/s2)a)40 m/sb)30 m/sc)27.5 m/sd)25 m/sCorrect answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
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