Question

Need help in showing how to get the final velocity​

Answer 1

To calculate the final velocity of an object, use the equation Vf = Vo + at, substituting in the known initial velocity, acceleration, and time.

Step-by-step explanation:

To calculate the final velocity (Vf) of an object in motion when its initial velocity (Vo), acceleration (a), and time (t) are known, you can use the basic kinematic equation:

v = vo + at

This equation comes from the definition of acceleration, which is the rate of change of velocity. Here's how you can use it:

1. Identify the known physical quantities: initial velocity (Vo), acceleration (a), and time (t).
2. Substitute these values into the kinematic equation to solve for the unknown final velocity (Vf).
3. Perform the calculation: Vf = Vo + (a × t).
4. Check if the calculated final velocity is reasonable in the context of the problem.

For example, if a car accelerates at 0.40 m/s² for 100 seconds from rest, its final velocity would be calculated as Vf = Vo + (a × t) = 0 + (0.40 m/s²) (100 s) = 40 m/s.

Answer 2

I don't think the provided solution is correct because it's not dimensionally consistent. The quantity
`8h-1` in particular doesn't make sense since it's mixing a distance with a dimensionless constant.

Here's how I think the proper answer should look:

The net force on the box acting perpendicular to the ramp is

`\sum F_\perp = F_(\rm normal) - mg \cos(\theta) = 0`

where
`F_(\rm normal)` is the magnitude of the normal force due to contact with the ramp and
`mg\cos(\theta)` is the magnitude of the box's weight acting in this direction. The net force is zero since the box doesn't move up or down relative to the plane of motion.

The net force acting parallel the ramp is

`\sum F_\| = mg\sin(\theta) - F_(\rm friction) = ma`

where
`mg\sin(\theta)` is the magnitude of the parallel component of the box's weight,
`F_(\rm friction)` is the magnitude of kinetic friction, and
`a` is the acceleration of the box.

From the first equation, we find

`F_(\rm normal) = mg \cos(\theta)`

and since
`F_(\rm friction) = \mu F_(\rm normal)`, we get from the second equation

`mg\sin(\theta) - \mu mg\cos(\theta) = ma`

and with
`\mu = 0.25` and
`\theta=60^\circ`, we get

`a = g\sin(60^\circ) - 0.25g \cos(60^\circ) = \left(\frac{\sqrt3}2 - \frac18\right) g`

Let
`x` be the length of the ramp, i.e. the distance that the box covers as it slides down it. Then the box attains a final velocity
`v` such that

`v^2 = 2ax`

From the diagram, we see that

`\sin(\theta) = \frac hx \implies x = \frac h{\sin(60^\circ)} = (2h)/(\sqrt3)`

and so

`v^2 = (4ah)/(\sqrt3) = \left(2 - \frac1{2\sqrt3}\right) hg = \frac14 \left(8 - \frac2{\sqrt3}\right) hg`

`\implies v = \frac12 \sqrt{\left(8 - \frac2{\sqrt3}\right) hg}`