Answer to Question #260067 in Physics for haze

Question

A block of mass 0.2 kg is pushed to be accelerated from rest to a speed of 4 m/s on a smooth surface. It then starts to climb up on a smooth slope. How high does the block climb on the slope until it stops in m (namely vertical height)? (Hint: Find first what the kinetic energy of the block is when its speed reaches 4 m/s from rest in J, and then calculate how high the block climbs on the slope until it stops in m). g=9.8 m/s²

Answer 1

1) Its kinetic energy:

$$KE=\frac12mv^2.$$

2) The kinetic energy is converted into potential energy:

$$KE=PE,\\\space\\ \frac12 mv^2=mgh,\\\space\\ h=\frac{v^2}{2g}=0.816\text{ m}.$$