Exam crakers 1001 physics question 184 185

[kg062008]

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If anyone has this book, I don't see how they got the answer for question 184,185, they just give the equation

d= 1/2at^2 at the end? How eactly did they find the time?😕

 

[Temperature101]

If anyone has this book, I don't see how they got the answer for question 184,185, they just give the equation

d= 1/2at^2 at the end? How eactly did they find the time?😕

F = ma ...Wsin30 = (W/g)a ....a = gsin30 ....a = 10*0.5 = 5 m/s^2


sin30 = opposite/hypotenuse.......sin30 = h/d ....0.5 = 5/d....d = 10


d=1/2at^2......10 = 1/2(5)t^2....t = 2s

 

[red doctober]

If anyone has this book, I don't see how they got the answer for question 184,185, they just give the equation

d= 1/2at^2 at the end? How eactly did they find the time?😕

b=angle of inclined plane above horizontal
sin(b)=h/d
sin(30)=5/d
d=5/sin(30)=5/0.5=10m

x is the axis parallel to the slope of the inclined plane
sin(b)=F_gx/F_g
Because the mass doesn't change F_gx/F_g=a_gx/a_g
a_g=10 m/s^2
From now for sake of simplicity on let's call a_gx "a"
sin(30)=a/10
a=10*sin(30)=10*0.5=5
d=0.5*at^2
10=0.5*5*t^2
t^2=4
t=2s

You can see that this is all independent of mass, so the answer for the next question is also 2s. Thus, it doesn't matter if it's a 1g mass or a 10-ton mass, it will always take the same amount of time (given that friction isn't an issue.) This principle can also be extended to a pendulum (period time is independent of bob mass.)