A wet bar of soap slides down a ramp 7.8m long inclined at 8.0 degrees?

Question

how long does it take to reach the bottom? 
assume mu=.052

Answer 1

A wet bar of a soap slides down from a ramp .

Ramp is Inclined at angle of 8° .

Length of ramp is 7.8 m .

Acceleration due to gravity g = 9.8

Coefficient of friction μ = 0.052 .

The net force F = mg sin ϴ - μ mg cos ϴ

F = ma = mg sin ϴ - μ mg cos ϴ

 ma = m (g sin ϴ - μ g cos ϴ )

a = g sin ϴ - μ g cos ϴ

So the acceleration of the soap is a = g sin ϴ - μ g cos ϴ .

a = (9.8) sin 8° - (0.052) (9.8) cos 8° 

a = 1.3638 - 0.5046 

a = 0.859 m/s² 

So the acceleration of the soap is 0.859  m/s² .

The body starts from rest so the initial velocity of body is zero .

Consider equation of motion .

V_f² - V_i² = 2as 

V_f² - 0² = 2(0.859)(7.8)

V_f²  = 13.4

V_f  = 3.66 m/s .

The time taken for the soap to reach the base is (distance moved ) / ( velocity of soap )

                        Time = 7.8 / 3.66

                       Time =  2.13 sec

So the time taken for the soap to move a distance 7.8 m is 2.13 sec .  

Comments

you are wrong.

---

The first part about finding the acceleration is correct but once you find the acceleration use the equation  (1/2)at^2, rearrange it for t. t= sqt. 2*x/a, where x is the length of the ramp. Then you can find the correct time.