A 2.00-kg block is pushed against a spring with negligible mass and force constant k =400N/m, compressing it 0.220 m. When the block is released, it moves along a frictionless, horizontal surface and then up a frictionless incline with slope 37.0° (Fig. P7.40). (a)What is the speed of the block as it slides along the horizontal surface after having left the spring? (b) How far does the block travel up the incline before starting to slide back down?

The given data can be listed below,

• The mass of the block is m =2 kg
• The force constant of the spring is, $k=400\mathrm{N}/\mathrm{m}$
• The distance spring is compressed is,$x=0.200\mathrm{m}$
• The inclination angle is,$\theta =37°$

Spring constant is a characteristic feature which is essentially a constant value for a particular material used to make the spring. This constant value gives the measure of stiffness of the spring when put under force or when any load is suspended.

The elastic potential energy of the block is given by,

$U_{el}=\frac{1}{2}k{x}^2$

Here, k is the force constant, and x is the compression.

For k =400 N/m, x =0.220 m , the above equation becomes-

$$\begin{gathered} U_{el}=\frac{1}{2}\left(400N/m\right){\left(0.22m\right)}^2 \\ =9.68\mathrm{J} \end{gathered}$$

This elastic potential energy is given to the block in the form of kinetic energy, when it is released

$K=\frac{1}{2}m{v}^2$

Here, m is the mass of the block and v is the velocity of the block.

From law of conservation of energy the velocity of the block is given by,

$\frac{1}{2}m{v}^2=9.68\mathrm{J}$

For m=2.0 kg, the velocity can be obtained as-

$$\begin{gathered} \mathrm{v}=\sqrt{\frac{2\times 9.68\mathrm{J}}{2\mathrm{kg}}} \\ =3.11\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the speed of the block is 3.11 m/s.

The velocity with which the body slides up is denoted by v whose value is 3.11 m/s.

As, the block starts gaining height the kinetic energy of the block will get converted into gravitational potential energy and at the highest point kinetic energy will be zero and gravitational potential energy will be maximum. Let the block reaches a maximum height h, then law of conservation of energy-

$$\begin{gathered} K=mgh \\ 9.68J=\left(2.0kg\right)\left(9.8m/s^2\right)h \\ h=\frac{9.68J}{\left(2.0kg\right)\left(9.8m/s^2\right)} \\ h=0.49m \end{gathered}$$

Let the distance travelled by block up the incline plane be d. then,

$$\begin{gathered} d\sin 37°=h \\ d=\frac{h}{\sin 37°} \\ d=\frac{0.49m}{\sin 37°} \\ d=0.82m \end{gathered}$$

Thus the distance the block travels upward is 0.82 m.