Question

A 75 kg. block of wood rest on the top of the smooth plane surface, whose length is 4 meters and whose altitude is 1.0 meter. How long will it take for th block to slide down to the bottom of the plane when released?

Answer 1

The speed of the block at the beginning of motion is $v_i =0m/s$. Ssnce the surface is smooth, the speed at the end of the motion $v_f$ can be find from the energy conservation law:

$$mgh=\dfrac{mv_f^2}{2}$$

where $m= 75kg, h = 1.0m, g= 9.8m/s^2$. Thus, obtain:

$$v_f = \sqrt{2gh}$$

Next, using the kinematic formula, find the time of motion $t$ :

$$d = \dfrac{v_i+v_f}{2}\cdot t$$

where $d = 4m$. Thus, obtain:

$$t = \dfrac{2d}{v_f} = \dfrac{2d}{\sqrt{2gh}} = d\sqrt{\dfrac{2}{gh}}\\ t = 4\cdot \sqrt{\dfrac{2}{9.8\cdot1}} \approx 2s$$

Answer. 2s.