Tracy pushes a block up an inclined plane to a height of 1.0 m.
If Tracy puts 475 J of work into the system, but only 450 J are used to elevate the block, what is the efficiency of the inclined plane?
A. 95%
B. 106%
c. 925)
OD 25 J

A=95%

(450/475)*100=94.7% which rounds to 95%

Initial velocity,  u = 4.9 m/sfinal velocity = v m/sinitial height,  h₁ =  245 m    final height, h₂ = 0a = -9.8  m/s²time taken = t secondss = ut + 0.5at²⇒  (h₂-h₁) = ut + 0.5at²⇒  0-245 = 4.9t + 0.5×(-9.8)×t²⇒ -245 = 4.9t - 4.9t²⇒ 4.9t² -4.9t -245 =0solving it, we get   t = 7.59sv  = u +  at = 4.9 -9.8×7.59 = 4.9 - 74.38 = -69.48 m/svelocity is  69.48 m/s downward