A 11.7-kg uniform board is wedged into a corner and held by a spring at a 50.0° angle, as the drawing shows. The spring has a spring constant of 230 N/m and is parallel to the floor. Find the amount by which the spring is stretched from its unstrained length.
To find the amount by which the spring is stretched, we will need to analyze the forces acting on the board.
We have the weight of the board acting vertically downward. The weight can be calculated using the formula:
weight = mass x acceleration due to gravity
Given that the mass of the board is 11.7 kg and the acceleration due to gravity is 9.8 m/s^2, the weight can be calculated as follows:
weight = 11.7 kg x 9.8 m/s^2 = 114.66 N
Now, we need to break down the weight into its vertical and horizontal components. The vertical component will oppose the force exerted by the spring and will be the force that stretches the spring.
The vertical component of the weight can be calculated using the formula:
vertical component = weight x sin(angle)
Given that the angle is 50.0°, the vertical component can be calculated as follows:
vertical component = 114.66 N x sin(50.0°) ≈ 87.63 N
Since the spring is parallel to the floor, the force exerted by the spring will also be vertical. The force exerted by the spring is given by Hooke's law:
force exerted by spring = spring constant x stretch
where the stretch is the amount by which the spring is stretched from its unstrained length.
Equating the force exerted by the spring to the vertical component of the weight, we can solve for the stretch:
spring constant x stretch = vertical component
Given that the spring constant is 230 N/m, the equation becomes:
230 N/m x stretch = 87.63 N
Solving for the stretch:
stretch = 87.63 N / 230 N/m ≈ 0.381 m
Therefore, the amount by which the spring is stretched from its unstrained length is approximately 0.381 m.
To find the amount by which the spring is stretched, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position.
The force exerted by the spring can be calculated using the equation:
F = k * x
Where:
- F is the force exerted by the spring (in newtons)
- k is the spring constant (in newtons per meter)
- x is the displacement from the equilibrium position (in meters)
In this case, we need to find the displacement, so we rearrange the equation to solve for x:
x = F / k
Now, let's calculate the force exerted by the spring:
First, let's find the vertical component of the weight of the uniform board. We can use the formula:
Weight = mass * gravitational acceleration
or
Weight = m * g
Here, m is the mass of the board and g is the gravitational acceleration. Assuming the standard value of 9.8 m/s^2 for g, we get:
Weight = 11.7 kg * 9.8 m/s^2 = 114.66 N
Now, let's find the component of the weight that is responsible for stretching the spring. This component is given by:
Force = Weight * sin(theta)
Using the given angle of 50.0°, we calculate:
Force = 114.66 N * sin(50.0°) = 87.71 N
Finally, we can find the displacement of the spring:
x = 87.71 N / 230 N/m ≈ 0.381 m
Therefore, the spring is stretched by approximately 0.381 meters from its unstrained length.