Question #260067

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




1
Expert's answer
2021-11-02T10:00:28-0400

1) Its kinetic energy:


KE=12mv2.KE=\frac12mv^2.

2) The kinetic energy is converted into potential energy:


KE=PE, 12mv2=mgh, h=v22g=0.816 m.KE=PE,\\\space\\ \frac12 mv^2=mgh,\\\space\\ h=\frac{v^2}{2g}=0.816\text{ m}.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS