Asked by jack
A block of mass M is projected up a frictionless inclined plane with a speed VO. The angle of incline is given as theta.
a. How far up the plane does it go?
b. How much time does it take to get there?
I was stuck after Fg(cos(theta+90)= Ma_x
gcos(90+theta)/M = a_x
can someone please help me
a. How far up the plane does it go?
b. How much time does it take to get there?
I was stuck after Fg(cos(theta+90)= Ma_x
gcos(90+theta)/M = a_x
can someone please help me
Answers
Answered by
Damon
kinetic energy at bottom = (1/2) M Vo^2
= potential energy at to when it stops
= m g h = M g (x sin theta)
so
(1/2) M Vo^2 = M g x sin thets
x = Vo^2/(2 g sin theta)
average speed = Vo/2
time = distance/average speed
= Vo^2/(2 g sin theta)] / [ Vo/2)]
= Vo/(g sin theta)
= potential energy at to when it stops
= m g h = M g (x sin theta)
so
(1/2) M Vo^2 = M g x sin thets
x = Vo^2/(2 g sin theta)
average speed = Vo/2
time = distance/average speed
= Vo^2/(2 g sin theta)] / [ Vo/2)]
= Vo/(g sin theta)
Answered by
jack
thanks but my instrutor don't want me to use energy to solve the problem since we havnt learn it yet, can you do it using newton's law?
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