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Q. A 2 kg block is dropped from a height of 0.4m on a spring of force constant k=1960nm^-1.the maximum compression of the spring is?
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Q. A 2 kg block is dropped from a height of 0.4m on a spring of force ...
Mass of the body is ( m ) = 2 Kg
spring's constant ( K ) = 1960 N/m
height of the body = 0.4 m
Let the maximum distance spring will compressed = x m
see the attachment , it is clear that
initial potential energy of body = mg(h + x)
final potential energy = spring potential energy = 1/2 Kx^2
according to law of conservation of energy,
initial potential energy = final potential energy
mg(h + x) = 1/2 Kx^2
2mgh + 2mgx = Kx^2
Kx^2 - 2mgx - 2mgh = 0
⇒ 1960x^2 - 2*2*9.8*x - 2*2*9.8*0.4 = 0
⇒ 1960x^2 - 39.2x - 15.68 = 0
after solving this quadratic equations , we get x = 0.1m
hence, answer is x = 0.1m
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Q. A 2 kg block is dropped from a height of 0.4m on a spring of force ...
Problem: A 2 kg block is dropped from a height of 0.4m on a spring of force constant k=1960 N/m. The maximum compression of the spring is?
Solution:
Given data:
- Mass of the block (m) = 2 kg
- Height (h) = 0.4 m
- Force constant (k) = 1960 N/m
Step 1: Calculate the potential energy of the block
The potential energy (PE) of the block is given by the formula:
PE = mgh
where m is the mass, g is the acceleration due to gravity, and h is the height.
Given: m = 2 kg, g = 9.8 m/s², h = 0.4 m
PE = (2 kg) x (9.8 m/s²) x (0.4 m)
= 7.84 J
Step 2: Calculate the maximum potential energy converted to spring potential energy
The maximum potential energy converted to spring potential energy is given by the formula:
PE_spring = (1/2) k x (x²)
where k is the force constant of the spring and x is the maximum compression of the spring.
Given: k = 1960 N/m
We need to find x.
Step 3: Equating the potential energy of the block to the spring potential energy
Since the potential energy of the block is converted to spring potential energy, we can equate the two equations:
PE = PE_spring
7.84 J = (1/2) (1960 N/m) (x²)
Simplifying:
7.84 J = 980 N/m (x²)
Dividing both sides by 980 N/m:
0.008 = x²
Taking the square root of both sides:
x = √(0.008) m
= 0.089 m
Therefore, the maximum compression of the spring is 0.089 m.
Solution:
Given data:
- Mass of the block (m) = 2 kg
- Height (h) = 0.4 m
- Force constant (k) = 1960 N/m
Step 1: Calculate the potential energy of the block
The potential energy (PE) of the block is given by the formula:
PE = mgh
where m is the mass, g is the acceleration due to gravity, and h is the height.
Given: m = 2 kg, g = 9.8 m/s², h = 0.4 m
PE = (2 kg) x (9.8 m/s²) x (0.4 m)
= 7.84 J
Step 2: Calculate the maximum potential energy converted to spring potential energy
The maximum potential energy converted to spring potential energy is given by the formula:
PE_spring = (1/2) k x (x²)
where k is the force constant of the spring and x is the maximum compression of the spring.
Given: k = 1960 N/m
We need to find x.
Step 3: Equating the potential energy of the block to the spring potential energy
Since the potential energy of the block is converted to spring potential energy, we can equate the two equations:
PE = PE_spring
7.84 J = (1/2) (1960 N/m) (x²)
Simplifying:
7.84 J = 980 N/m (x²)
Dividing both sides by 980 N/m:
0.008 = x²
Taking the square root of both sides:
x = √(0.008) m
= 0.089 m
Therefore, the maximum compression of the spring is 0.089 m.
Community Answer
Q. A 2 kg block is dropped from a height of 0.4m on a spring of force ...
9.1 cm
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Q. A 2 kg block is dropped from a height of 0.4m on a spring of force constant k=1960nm^-1.the maximum compression of the spring is?
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Q. A 2 kg block is dropped from a height of 0.4m on a spring of force constant k=1960nm^-1.the maximum compression of the spring is? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Q. A 2 kg block is dropped from a height of 0.4m on a spring of force constant k=1960nm^-1.the maximum compression of the spring is? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Q. A 2 kg block is dropped from a height of 0.4m on a spring of force constant k=1960nm^-1.the maximum compression of the spring is?.
Q. A 2 kg block is dropped from a height of 0.4m on a spring of force constant k=1960nm^-1.the maximum compression of the spring is? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Q. A 2 kg block is dropped from a height of 0.4m on a spring of force constant k=1960nm^-1.the maximum compression of the spring is? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Q. A 2 kg block is dropped from a height of 0.4m on a spring of force constant k=1960nm^-1.the maximum compression of the spring is?.
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