speed of roller coaster... how to work problem?

Question

suppose a roller coaseter passes point one (summit of a hill) at a height of 35m amd a speed pf 1.70m/s. if the average force of frictions is equal to one-fifth of its weight, what speed will it reach point 2 (the bottom of the hill) at 0m? the distance traveled is 45.0m.

Answer 1

You know that for the most part, energy is conserved:

Eo = Ef

However, energy may be lost due to friction (We is the external work done by friction):

mgh + (1/2)mv² - We= mgh + (1/2)mv²

Substitute in the formula for work, and plug in the values:

mgh + (1/2)mv² - (1/5)(45.0)mg= mgh + (1/2)mv²

Now factor out the "m" and cancel it from both sides:

gh + (1/2)v² - (1/5)(45.0)g= gh + (1/2)v²

Now just plug in values and simplify:

(9.80)(35) + (1/2)(1.70)² - (1/5)(45.0)(9.80) = (9.80)(0) + (1/2)v²

3.4x10^2 + 1.45 - 88.2 = (1/2)v²

(1/2)v² = 253

v² = 506

v = 22.5 m/s

Answer 2

Since the force of friction is 1/5 of the weight, the roller coaster would only acquire 4/5 of the energy it would without the friction. This energy is .5*m*v^2=.5*m*(1.7)^2+.8*m*g*45
or, v^2=1.7^2+1.6g*45=708.49
v=26.61m/s