A small block with mass 0.200 kg is released from rest at the top of a frictionless incline. The block travels a distance 0.394 m down the incline in 2.00 s. The 0.200 kg block is replaced by a 0.400 kg block. If the 0.400 kg block is released from rest at the top of the incline, how far down the incline does it travel in 2.00 s

The distance traveled by the second block is 0.197 m

Explanation:

Given;

mass of the small block, m₁ = 0.2 kg

distance traveled by the block, d₁ =0.394 m

time of travel, t = 2 s

mass of the second block, m₂ = 0.4 kg

distance traveled by the second block, d₂ = ?

The work done per unit time on the inclined plane is given by;

Therefore, the distance traveled by the second block is 0.197 m

b

step-by-step explanation: a third of 24 is 8, 8 times 3 is 24 so 8 - 2 would be 6