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A tank of horizontal cross-sectional area 1m2, a spring with force constant 2000 Nm1 is fixed in vertical position upto the height of the water as shown in figure 1. A block of mass 180 kg is gently placed over the spring and it attains the equilibrium position as shown in figure 2. If base area of the block is 0.2m2 and height 60 cm, then find compression in the spring in cm in equilibrium position. (take g 10 m/s p 1000 kg/m3).?
Most Upvoted Answer
A tank of horizontal cross-sectional area 1m2, a spring with force con...
Find the Compression in the Spring in Equilibrium Position:
Given:
- Horizontal cross-sectional area of the tank: 1 m^2
- Force constant of the spring: 2000 N/m
- Mass of the block: 180 kg
- Base area of the block: 0.2 m^2
- Height of the block: 60 cm
- Acceleration due to gravity: 10 m/s^2
- Density of water: 1000 kg/m^3
Let's break down the problem into different steps:
Step 1: Calculate the weight of the block
The weight of the block can be calculated using the formula:
Weight = mass * gravitational acceleration
Weight = 180 kg * 10 m/s^2 = 1800 N
Step 2: Calculate the buoyant force on the block
The buoyant force on the block can be calculated using the formula:
Buoyant force = density of fluid * volume of fluid displaced * gravitational acceleration
Since the block is fully submerged in water, the volume of fluid displaced is equal to the volume of the block.
Volume of the block = base area of the block * height of the block
Volume of the block = 0.2 m^2 * 0.6 m = 0.12 m^3
Buoyant force = 1000 kg/m^3 * 0.12 m^3 * 10 m/s^2 = 1200 N
Step 3: Calculate the net force acting on the block
The net force acting on the block can be calculated by subtracting the buoyant force from the weight of the block.
Net force = Weight - Buoyant force
Net force = 1800 N - 1200 N = 600 N
Step 4: Calculate the compression in the spring
The compression in the spring can be calculated using Hooke's Law, which states that the force exerted by a spring is directly proportional to its compression or extension.
F = k * x
Where F is the force exerted by the spring, k is the force constant, and x is the compression or extension.
In equilibrium, the net force acting on the block is equal to the force exerted by the spring.
Net force = Force exerted by the spring
600 N = 2000 N/m * x
x = 600 N / 2000 N/m = 0.3 m
Step 5: Convert the compression to centimeters
The compression in meters can be converted to centimeters by multiplying by 100.
Compression in cm = 0.3 m * 100 cm/m = 30 cm
Therefore, the compression in the spring in the equilibrium position is 30 cm.
Given:
- Horizontal cross-sectional area of the tank: 1 m^2
- Force constant of the spring: 2000 N/m
- Mass of the block: 180 kg
- Base area of the block: 0.2 m^2
- Height of the block: 60 cm
- Acceleration due to gravity: 10 m/s^2
- Density of water: 1000 kg/m^3
Let's break down the problem into different steps:
Step 1: Calculate the weight of the block
The weight of the block can be calculated using the formula:
Weight = mass * gravitational acceleration
Weight = 180 kg * 10 m/s^2 = 1800 N
Step 2: Calculate the buoyant force on the block
The buoyant force on the block can be calculated using the formula:
Buoyant force = density of fluid * volume of fluid displaced * gravitational acceleration
Since the block is fully submerged in water, the volume of fluid displaced is equal to the volume of the block.
Volume of the block = base area of the block * height of the block
Volume of the block = 0.2 m^2 * 0.6 m = 0.12 m^3
Buoyant force = 1000 kg/m^3 * 0.12 m^3 * 10 m/s^2 = 1200 N
Step 3: Calculate the net force acting on the block
The net force acting on the block can be calculated by subtracting the buoyant force from the weight of the block.
Net force = Weight - Buoyant force
Net force = 1800 N - 1200 N = 600 N
Step 4: Calculate the compression in the spring
The compression in the spring can be calculated using Hooke's Law, which states that the force exerted by a spring is directly proportional to its compression or extension.
F = k * x
Where F is the force exerted by the spring, k is the force constant, and x is the compression or extension.
In equilibrium, the net force acting on the block is equal to the force exerted by the spring.
Net force = Force exerted by the spring
600 N = 2000 N/m * x
x = 600 N / 2000 N/m = 0.3 m
Step 5: Convert the compression to centimeters
The compression in meters can be converted to centimeters by multiplying by 100.
Compression in cm = 0.3 m * 100 cm/m = 30 cm
Therefore, the compression in the spring in the equilibrium position is 30 cm.
Community Answer
A tank of horizontal cross-sectional area 1m2, a spring with force con...
Is a answer 40 m
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A tank of horizontal cross-sectional area 1m2, a spring with force constant 2000 Nm1 is fixed in vertical position upto the height of the water as shown in figure 1. A block of mass 180 kg is gently placed over the spring and it attains the equilibrium position as shown in figure 2. If base area of the block is 0.2m2 and height 60 cm, then find compression in the spring in cm in equilibrium position. (take g 10 m/s p 1000 kg/m3).?
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A tank of horizontal cross-sectional area 1m2, a spring with force constant 2000 Nm1 is fixed in vertical position upto the height of the water as shown in figure 1. A block of mass 180 kg is gently placed over the spring and it attains the equilibrium position as shown in figure 2. If base area of the block is 0.2m2 and height 60 cm, then find compression in the spring in cm in equilibrium position. (take g 10 m/s p 1000 kg/m3).? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A tank of horizontal cross-sectional area 1m2, a spring with force constant 2000 Nm1 is fixed in vertical position upto the height of the water as shown in figure 1. A block of mass 180 kg is gently placed over the spring and it attains the equilibrium position as shown in figure 2. If base area of the block is 0.2m2 and height 60 cm, then find compression in the spring in cm in equilibrium position. (take g 10 m/s p 1000 kg/m3).? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A tank of horizontal cross-sectional area 1m2, a spring with force constant 2000 Nm1 is fixed in vertical position upto the height of the water as shown in figure 1. A block of mass 180 kg is gently placed over the spring and it attains the equilibrium position as shown in figure 2. If base area of the block is 0.2m2 and height 60 cm, then find compression in the spring in cm in equilibrium position. (take g 10 m/s p 1000 kg/m3).?.
A tank of horizontal cross-sectional area 1m2, a spring with force constant 2000 Nm1 is fixed in vertical position upto the height of the water as shown in figure 1. A block of mass 180 kg is gently placed over the spring and it attains the equilibrium position as shown in figure 2. If base area of the block is 0.2m2 and height 60 cm, then find compression in the spring in cm in equilibrium position. (take g 10 m/s p 1000 kg/m3).? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A tank of horizontal cross-sectional area 1m2, a spring with force constant 2000 Nm1 is fixed in vertical position upto the height of the water as shown in figure 1. A block of mass 180 kg is gently placed over the spring and it attains the equilibrium position as shown in figure 2. If base area of the block is 0.2m2 and height 60 cm, then find compression in the spring in cm in equilibrium position. (take g 10 m/s p 1000 kg/m3).? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A tank of horizontal cross-sectional area 1m2, a spring with force constant 2000 Nm1 is fixed in vertical position upto the height of the water as shown in figure 1. A block of mass 180 kg is gently placed over the spring and it attains the equilibrium position as shown in figure 2. If base area of the block is 0.2m2 and height 60 cm, then find compression in the spring in cm in equilibrium position. (take g 10 m/s p 1000 kg/m3).?.
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