A crate of mass M starts from rest at the top of a frictionless ramp inclined at an angle α above the horizontal. Find its speed at the bottom of the ramp, a distance d from where it started. Do this in two ways.

The speed of the crate at the bottom of the ramp is

The given parameters;

The normal force on the crate is calculated as follows;

Fₙ = mgcosα

The net horizontal force on the crate is calculated as follows;

mgsinα = ma

gsinα = a

The speed of the crate at the bottom of the ramp is calculated as follows;

Thus, the speed of the crate at the bottom of the ramp is

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Complete question is;

A crate of mass M starts from rest at the top of a frictionless ramp inclined at an angle α above the horizontal. Find its speed at the bottom of the ramp, a distance d from where it started. Do this in two ways: (a) Take the level at which the potential energy is zero to be at the bottom of the ramp with y positive upward. (b) Take the zero level for potential energy to be at the top of the ramp with y positive upward. (c) Why did the normal force not enter into your solution?

A) v2 = √(2g•d•sin α)

B) v2 = √(2g•d•sin α)

C) Because normal force is perpendicular to the displacement along the ramp and thus the work done is zero

Explanation:

Mass of the crate = M

Angle of ramp above horizontal = α

Distance between starting point to end point = d

A) Since, we have d and α, then the height of the ramp will be found using trigonometric ratios for a right angle triangle.

Thus;

h = d sin α

When the potential energy at the bottom of the ramp is zero, we have;

y1 = h and y2 = 0

Now, we know that total work done is given by the formula;

W_tot = K2 - K1 = W_other + W_grav

But since gravitational force is the only force acting on the crate in the direction of motion, then W_other = 0.

Now, W_grav is given by the formula;

W_grav = mgy1 - mgy2

So, we now have;

K2 - K1 = mgy1 - mgy2

Where K2 and K1 are final and initial kinetic energy respectively.

So,

½m(v2)² - ½m(v1)² = mgy1 - mgy2

Since v1 and y2 are zero, and y1 = h, then we have;

½m(v2)² = mgh

Cross multiply to get;

(v2)² = 2gh

We earlier established that h = d sin α

So,

(v2)² = 2g•d•sin α

v2 = √(2g•d•sin α)

B) when the potential energy is zero at the top of the tamp, we have;

y1 = 0 and y2 = -h

Like in solution A above, W_other = 0

Similarly,

½m(v2)² - ½m(v1)² = mgy1 - mgy2

We have, v1 and y1 are zero, and y2 = - h, then we have;

½m(v2)² = -mg(-h)

Cross multiply to get

(v2)² = 2gh

We earlier established that h = d sin α

So,

(v2)² = 2g•d•sin α

v2 = √(2g•d•sin α)

C) I did not enter the normal force into the solution because the normal force is perpendicular to the displacement along the ramp and thus the work done is zero

nec chapter 5 contains

c. information on special occupancy areas

nec chapter 6 covers

a. special equipment

which of the following rules contain the words shall or shall not and must be adhered to?

b. mandatory rules

which of the following rules identify actions that are allowed but not required and typically cover options or alternative methods?

c. optional rules

which of the following serves as the introduction to the nec and provides an overview of the book?

a. nec article 90

organizations that test products for safety are

c. nrtls

the nec is published by the

a national fire protection association

the nec is divided into

b. 9 chapters

the nec provides informative annexes for

c. informational materials and examples

two terms from nec article 100 that trainees should become familiar with are

b. labeled and listed

explanation: