Question

A block is projected up a frictionless inclined plane with initial speed v0 = 7.14 m/s. The angle of incline is θ = 36.5°. (a) How far up the plane does it go? (b) How long does it take to get there? (c) What is its speed when it gets back to the bottom?

Answer 1

Answer:

$(a)x=4.37m\\\\(b)t=1.225s\\\\(c) v_{f}=7.14m/s$

Explanation:

Given data

$v_{o}=7.14m/s\\\alpha =36.5^o$

For Part (a)

Starting with the -ve acceleration of the body (opposite to the gravitational force)

$a=-gSin\alpha \\a=-(9.8m/s^2)Sin(36.5)\\a=-5.83m/s^2$

Using equation of motion

$v_{f}^2=v_{o}^2+2ax\\(0m/s)^2=(7.14m/s)^2+2(-5.83m/s^2)x\\-(7.14m/s)^2=2(-5.83m/s^2)x\\x=\frac{-(7.14m/s)^2}{2(-5.83m/s^2}\\ x=4.37m$

For Part (b)

Using the result in Part (a) we can substitute in other equation of motion to get time t:

$x=\frac{1}{2}vt\\ 4.37m=\frac{1}{2}(7.14m/s)t\\ (7.14m/s)t=2*(4.37)\\t=8.744/7.14\\t=1.225s$

For Part (c)

At state 2 where vo=0m/s and the acceleration is positive (same direction as the gravitational force)

$a=gSin\alpha \\a=(9.8m/s^2)Sin(36.5)\\a=5.83m/s^2\\\\\\v_{f}^2=v_{o}^2+2ax\\v_{f}^2=(0m/s)^2+2(5.83m/s^2)(4.37m)\\v_{f}^2=50.95\\v_{f}=\sqrt{50.95}\\ v_{f}=7.14m/s$