A block is released from rest at the top of a hill of height h. If there is negligible friction between the block and the hill, the block arrives at the bottom of the hill with speed v.

v / √2

Explanation:

A blocked from rest at the top of a hill of height h it there is negligible friction between the block and the hill, the block arrives at the bottom of the hill with speed. The block is released from rest at the top of another hill with a rough surface and height h. if one-half of the initial mechanical energy of the block Earth System is lost due to friction as the block descends the hill, the block will reach the bottom of the hill with a speed of?

Solution:

For the first block block, its potential energy at the top of the hill when it is at rest is converted to kinetic energy after release with a velocity v.

The kinetic energy = (1/2)mv², where m is the mass of the block.

For the other block placed on the top of the rough hill, the mechanical energy (kinetic energy) is halved due to the roughness of the hill. If u  is the speed of this block at the bottom of the hill then:

kinetic energy of block on the rough hill = (1//2)mu²

Hence:

(1//2)mu² = half of main system kinetic energy

(1//2)mu² = 1/2 * (1/2)mv²

mu² = (1/2)mv²

u² = (1/2)v²

u = √(v²/2)

u = v / √2

Hence the speed at the bottom of the rock of the block placed on the rough surface is v / √2

f=ma a/f/m a= 480/60 a = 8m/s/s that is quick wow