Question

A block is projected up a frictionless inclined plane with initial speed v0 = 3.50 m/s. The angle of incline is θ =32.0°. (a) How far up the plane does the block go? (b) How long does it take to get there? (c) What is its speedwhen it gets back to the bottom?

Answer 1

a). 1.18 m

b). 0.67 s

c). 3.5
`(m)/(s)`

Step-by-step explanation:

`a=-g*sen*(32)= -5.1932 (m)/(s^(2) )= -5.2 (m)/(s^(2) )`

a).

`V_(f) ^(2) =V_(i) ^(2)+2*a*d` ,
`V_(f) =0`,
`V_(i) =3.5(m)/(s )`

`d= (3.5^(2)(m)/(s)  )/(2*5.2(m)/(s^(2) ) ) = 1.179426 m\\d=1.18m`

b).

`V_(f) = V_(i) + a *t` ,
`V_(f) =0`,
`V_(i) =3.5(m)/(s )`

`0 = 3.5 + 5.2*t`

`t= (3.5 (m)/(s) )/(5.2 (m)/(s^(2) ) )`

`t= 0.67 s`

c).

`V_(f) ^(2) =V_(i) ^(2)+2*a*d`,
`V_(i) =0`

`V_(f) ^(2) = 0 + 2* 5.2 (m)/(s^(2) ) * 1.18m = 12.272 (m^(2) )/(s^(2) )`

`V_(f) =\sqrt{12.272 (m^(2) )/(s^(2) ) } =3.503 (m)/(s)`