- #1

- 287

- 46

- Homework Statement
- Attached image below!

- Relevant Equations
- Conservation of energy, kinematic equations

d) The block of mass m is pushed toward the wall until the spring has been compressed, and so the spring has stored PE, and when the block is released, it follows a projectile motion and strikes the floor ##stored PE\rightarrow K.E ##.

By law of conservation of energy,

$$ E_i = E_f $$

$$ mgh + \frac{1}{2}kx^2 = \frac {1}{2}mv_f^2$$

$$ k = \frac{2m}{x^2} (\frac {v^2}{2} - gh)$$

a) The time elapsed from the instant the block leaves the table to the instant it strikes the floor is taken as ##\Delta t##.

$$ E_i = E_f $$

$$ mgh + \frac{1}{2}kx^2 = \frac {1}{2}mv_f^2$$

I replace ##v_f## in the above equation with range D over elapsed time.

$$mgh + \frac{1}{2}kx^2 = \frac{1}{2}m \frac{D^2}{\Delta t^2}$$

In the above equation I substitute for k which I got from part d.

$$\Delta t = \sqrt {\frac {D}{V}}$$

b) When the block takes a projectile motion, it has uniform motion at constant velocity in the horizontal direction.

There fore ## (v_x)_f = (v_x)_i## = constant

c) The work done on the block by the spring is given by

$$ W.D_{sp} = \frac{1}{2}kx^2$$

Substituting k in the above equation we get,

$$ W.D_{sp} = m (\frac {v^2}{2} - gh)$$

Have I done it right?

By law of conservation of energy,

$$ E_i = E_f $$

$$ mgh + \frac{1}{2}kx^2 = \frac {1}{2}mv_f^2$$

$$ k = \frac{2m}{x^2} (\frac {v^2}{2} - gh)$$

a) The time elapsed from the instant the block leaves the table to the instant it strikes the floor is taken as ##\Delta t##.

$$ E_i = E_f $$

$$ mgh + \frac{1}{2}kx^2 = \frac {1}{2}mv_f^2$$

I replace ##v_f## in the above equation with range D over elapsed time.

$$mgh + \frac{1}{2}kx^2 = \frac{1}{2}m \frac{D^2}{\Delta t^2}$$

In the above equation I substitute for k which I got from part d.

$$\Delta t = \sqrt {\frac {D}{V}}$$

b) When the block takes a projectile motion, it has uniform motion at constant velocity in the horizontal direction.

There fore ## (v_x)_f = (v_x)_i## = constant

c) The work done on the block by the spring is given by

$$ W.D_{sp} = \frac{1}{2}kx^2$$

Substituting k in the above equation we get,

$$ W.D_{sp} = m (\frac {v^2}{2} - gh)$$

Have I done it right?