When A 3 kg object is suspeneded by an elastic band, it is stretched 15 cm. What would the srping constant for the elastic be?
k= weight/distance= 3*9.8/.15 N/m
If a force of 200 N accelerates a 20 kg sled at 8.0 m/s2, the friction force would be:
Why did the object bring a measuring tape to the elastic band party? Because it wanted to find out the "spring constant" and stretch the limits of their friendship! Ba-dum tss!
To calculate the spring constant (k) for the elastic band, we can use Hooke's Law, which states that the force exerted by an elastic object is proportional to the displacement of the object from its equilibrium position. The formula is F = -kx, where F is the force, k is the spring constant, and x is the displacement.
In this case, the force exerted by the object is its weight, which is given by the formula F = mg, where m is the mass of the object (3 kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2). As we know the displacement (15 cm), we can set up the equation as follows:
F = -kx
mg = -kx
3 kg * 9.8 m/s^2 = -k * 0.15 m
Solving for k, we get:
k = (3 kg * 9.8 m/s^2) / 0.15 m
Calculating this, we find that the spring constant (k) for the elastic band is approximately 196 N/m.
So, the elastic band is stretchy, but hopefully it won't "spring" any surprise parties on you!
To find the spring constant (k) for the elastic band, we can use Hooke's Law equation:
F = k * x
where:
F is the force applied on the elastic band
k is the spring constant
x is the displacement/stretch of the elastic band
Given information:
Mass of the object (m) = 3 kg
Displacement/stretch of the elastic band (x) = 15 cm = 0.15 m
Now we need to find the force applied on the elastic band. The force can be calculated using Newton's Second Law of Motion:
F = m * g
where:
m is the mass
g is the acceleration due to gravity (= 9.8 m/s^2)
Substituting the given values:
F = 3 kg * 9.8 m/s^2
F = 29.4 N
Now we can rearrange Hooke's Law equation to solve for the spring constant (k):
k = F / x
Substituting the values:
k = 29.4 N / 0.15 m
k ≈ 196 N/m
Therefore, the spring constant for the elastic band is approximately 196 N/m.
To find the spring constant for the elastic band, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the object from its equilibrium position.
The formula for Hooke's Law is F = -kx, where F is the force exerted by the spring, k is the spring constant, and x is the displacement of the object.
In this case, we know that the object has a mass of 3 kg and is stretched 15 cm (0.15 meters). The force exerted by the elastic band is due to the gravitational force acting on the object, which is given by the formula F = mg, where m is the mass of the object (3 kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Therefore, the force exerted by the elastic band is F = (3 kg) * (9.8 m/s^2) = 29.4 N.
Now we can solve for the spring constant k. Rearranging the formula F = -kx, we get k = -F/x.
Plugging in the values, we have k = -(29.4 N) / (0.15 m) ≈ -196 N/m.
Thus, the spring constant for the elastic band is approximately 196 N/m.