Hookes law states that the force , F in a spring extended by length x is given by F=-Kx. According to Newton's second law F=ma, where m is the mass and a is the acceleration. Calculate the dimension of the spring constant k.
-K x = m a
-K ⋅ meters = kg ⋅ m/s^2 = Newtons
K = - N / m
the spring force increases linearly (directly proportional to the stretch)
... in the direction opposite the elongation (stretching)
Well, let's analyze this situation like a true clown bot! According to Hooke's law, the force F in a spring is given by F = -Kx. According to Newton's second law, F = ma. So, we can equate these two to find our answer!
Equating the two equations, we have ma = -Kx. Now let's consider the dimensions. The dimension of mass (m) is represented by [M]. The dimension of acceleration (a) is represented by [L/T^2]. The dimension of force (F) is represented by [M.L/T^2]. And the dimension of length (x) is represented by [L].
Now, in the equation ma = -Kx, we can replace the dimensions:
[M] * [L/T^2] = -K * [L].
From this, we can isolate the dimension of the spring constant K:
[K] = [M] * [L/T^2] / [L].
Simplifying this, [K] = [M/T^2].
So, the dimension of the spring constant "K" is [M/T^2]. Ta-da!
To calculate the dimension of the spring constant (K), we can analyze the equation F = -Kx using dimensional analysis.
According to Hook's Law, the force (F) in a spring is directly proportional to the displacement (x) and is given by F = -Kx, where K is the spring constant.
Now let's compare the dimensions of the variables involved in the equation:
The dimension of force (F) is [M L T^-2], where M represents mass, L represents length, and T represents time.
The dimension of displacement (x) is [L], representing length.
The dimension of the spring constant (K) can be represented as [M L T^-2]/[L] or [M T^-2].
By equating the dimensions of both sides of the equation, we have:
[M L T^-2] = -[M T^-2] [L]
Simplifying and canceling the dimensions, we get:
[M] = [M]
The dimensions on both sides of the equation are the same, thus confirming that the dimension of K is [M T^-2].
To calculate the dimension of the spring constant, we need to consider the equation F = -Kx in the context of Newton's second law.
Newton's second law states that the force (F) acting on an object is equal to the product of its mass (m) and its acceleration (a): F = ma.
In the case of Hooke's law, the force (F) is given by F = -Kx, where K is the spring constant and x is the displacement of the spring from its equilibrium position.
Since both F = ma and F = -Kx describe the same force, we can equate these two equations:
ma = -Kx
The force constant K must have dimensions that make the equation consistent. By analyzing the dimensions on both sides of the equation, we can determine the dimension of K.
On the left side, the dimension of F is mass (M) times acceleration (L/T^2), which gives us [F] = ML/T^2.
On the right side, the dimension of K is force (M.L/T^2) divided by displacement (L). Therefore, the dimension of K is [K] = ML/T^2 / L = ML^2/T^2.
Hence, the dimension of the spring constant K is ML^2/T^2.