A spring ( K= 200 N/m) is fixed at the top of a frictionless plane inclined at angle $\mathit{\theta }\mathbf{=}\mathbf{40}\mathbf{°}$(Fig. 8-59). A 1.0 kgblock is projected up the plane, from an initial position that is distance d = 0.60 mfrom the end of the relaxed spring, with an initial kinetic energy of 16 J. (a) What is the kinetic energy of the block at the instant it has compressed the spring 0.20 m? (b) With what kinetic energy must the block be projected up the plane if it is to stop momentarily when it has compressed the spring by 0.40 m?

i) The spring constant is,$k=200\frac{N}{m}$

ii) The angle of inclination is,$\theta =40°$

iii) The mass of the block is,$m=1.0kg$

iv) The initial kinetic energy is$K_A=0.20m$

v) The spring compressed,$x_c=0.20m$

vi) The spring compressed,$x{\text{'}}_c=0.40m$

Here, using the conservation of energy we get, the total potential energy is the sum of gravitational potential energy and elastic potential energy of the spring. Also, using the equation for total potential energy U, we can find the kinetic energy ${\mathit{K}}_{\mathbf{c}}$of the block at the instant it has compressed the spring must the 0.20 mand kinetic energy$\mathit{K}{\mathbf{\text{'}}}_{\mathbf{A}}$of the block at the instant it has compressed the spring must the 0.40 m.

Formula:

1. The energy conservation is,$K_A+U_A=K_B+U_B=K_C+U_C$
2. The total potential energy is,$U=U_g+U_s=mgy+\frac{1}{2}k{x}^2$

As the block is projected up the incline. Its kinetic energy is converted into gravitational potential energy and elastic potential energy of the spring. The block compresses the spring, stopping momentarily before sliding back down again.

Let A be the starting point and the reference point for computing gravitational potential energy. That is,

$U_A=0$

The block comes into contact with the spring at B. The spring is compressed by an additional amount at C, as shown in the figure below.

By energy conservation, we get,

$$\begin{gathered} K_A+U_A=K_B+U_B \\ =K_C+U_C \end{gathered}$$

Here, we have to note that,

$$\begin{gathered} U=U_g+U_s \\ =mgy+\frac{1}{2}k{x}^2 \end{gathered}$$

That is, the total potential energy is the sum of gravitational potential energy and elastic potential energy of the spring.

At the instant when$x_c=0.20m$,

Solve for the vertical height as:

$$\begin{gathered} y_c=\left(d+x_c\right)\sin \theta \\ \Rightarrow y_c=\left(0.60+0.20\right)\sin 40 \\ \Rightarrow y_c=0.514m \end{gathered}$$

Applying the energy conservation principle, we get,

$K_A+K_A=K_C+U_C$

But,$U_C=mg{y}_c+\frac{1}{2}k{x}_c^2$putting, we get,

$\Rightarrow 16+0=K_c+mg{y}_c+\frac{1}{2}k{x}_c^2$

From this equation, we get,

$$\begin{gathered} K_A=K_c+mg{y}_c+\frac{1}{2}k{x}_c^2 \\ \Rightarrow K_c=16-1\times 9.8\times 0.514-\frac{1}{2}\times 200\times 0.{20}^2 \\ \Rightarrow K_c\approx 7.0J \end{gathered}$$

Similarly, at the instant when $x{\text{'}}_c=0.40m$ , the vertical height is given by,

$$\begin{gathered} y{\text{'}}_c=\left(d+x{\text{'}}_c\right)\sin \theta \\ \Rightarrow y{\text{'}}_c=\left(0.60+0.40\right)\sin 40 \\ \Rightarrow y{\text{'}}_c=0.64m \end{gathered}$$

Applying the energy conservation principle, we have

$K{\text{'}}_A+U{\text{'}}_A=K{\text{'}}_C+U{\text{'}}_C$

Since, $U{\text{'}}_A=0$, the initial kinetic energy that gives $K{\text{'}}_C=0$is

$$\begin{gathered} K{\text{'}}_A=U{\text{'}}_C \\ =mgy\text{'}+\frac{1}{2}kx{\text{'}}_c^2 \end{gathered}$$

Solve for the kinetic energy as:

$$\begin{gathered} \Rightarrow K{\text{'}}_A=\left(1.0\times 9.8\times 0.64\right)+\frac{1}{2}\left(200\times 0.{40}^2\right) \\ \Rightarrow K{\text{'}}_A=22J \end{gathered}$$