A woman pushes an initially stationary 11.0 kg gift up a 19.1 degree ramp a distance of 8.09 m.

Question

If she pushes with constant force of 89.7 N and the coefficient of friction is 0.399, what is the final velocity of the gift?

Answer 1

A woman pushes an initially stationary 11.0 kg gift up a 19.1 degree ramp a distance of 8.09 m.

If she pushes with constant force of 89.7 N and the coefficient of friction is 0.399, what is the final velocity of the gift?

Hello Jen.

The question above uses the topic of multiple forces upon on object while utilizing kinematic equations.

As we begin to solve this problem, it would be best if we work to find all forces acting upon the gift. Remember that each of these forces will play a factor and part in finding the acceleration of the box up the slope translating into the final velocity using kinematics.

As a check when you are finding the forces you should keep in mind...

Force of the woman on the gift.

Force of Friction on the gift

Force of Normal on the gift

Force of Gravity on the gift.

While drawing a force diagram, you will find something very peculiar to starting force diagrams.

This peculiarity may be seen though the use of the slope of the ground which may seem to shift everything into multiple complicated x and y portions.

I would like to propose a way for you to simplify your work while working with this system. Imagine shifting your paper 19.1 degrees so that you have the Force of the Woman and the Frictional force on the X-Axis while your Force of Normal is straight up on the Y axis. Now, the 19.1 degree shift has affected your Force of gravity changing it to hold both x and y components. I will later explain how you can use this to your advantage when solving for Force of Normal and Force of Friction. By shifting all of the paper you have acted upon every part of your equation therefore retaining the same final answer while avoiding mistakes with multiple x-y components.

Force of Gravity = Ma = (11kg)*(9.8m/s^2) = 107.8 kgm/s^2= 107.8N

Force of Normal (end up canceling out with the y component to Force of Gravity)

Newton's Second law of motion is often looked at a simply...

F = ma

What if two linebackers are dug into the ground while pushing one another? Both linebackers push each other with the same force, but they are not moving. This phenomenon can be explained thorough the fact that they are 'dug into the ground' by their positioning the linebackers use other forces such as their separate forces of normal on the ground.

When you think about it, separate forces are able to act on an object or system in order to maintain equilibrium or to create an acceleration in a direction.

Lets put this to use by applying Newton's second law in a different light,,,

The Summation of the Forces in the y direction = Force of Gravity(y-component) + Force of Normal

The Summation of the Forces in the y direction = Mass * acceleration

In real life have you ever seen anyone push a box or any object parallel to a surface without additional forces other than gravity and normal?

If your answer to that is no then you have found a key part that will prove to you that since the object doesn't start floating up into the sky at an increasing rate or sink into the surface at an increasing rate the acceleration is 0

Therefore, Force of Gravity (y-component) + Force of Normal = 11.0kg * 0

Since we know that the force of gravity is magnitude 107.8 we can create a right triangle with x and y components separated and the magnitude as the hypotenuse.

Cos(theta) = gravity(y-component) / Force of gravity magnitude

Force of gravity magnitude * Cos (theta) = gravity(y-component)

107.8cos(19.1) = 101.865 = gravity(y-component)

-101.865(force of gravity y)+ x(force of normal) = 0

Force of Normal = 101.868

Force of Gravity (x- component) will be found using sin

Sin(theta) = gravity(x-component) / Force of gravity magnitude

Force of gravity magnitude * Sin (theta) = gravity(x-component)

107.8sin(19.1) = 35.274 = gravity(x-component)

Lets make a check of where we are at right now...

Force of gravity magnitude =| -107.8| #note negative in my reference frame is downward for y

Force of gravity (y-component) magnitude=| -101.865| # negative in my reference frame is downward for y

Force of gravity (x-component) magnitude =| -35.274|

# negative in my reference frame is to the right (away from direction of force of woman on box

Force of Normal magnitude = |101.865| #positive in my reference frame is upward for y

Force of Woman on gift = 98.7N (given) #positive in x reference frame

Force of Friction on gift = Coefficient of friction * Force of Normal = 101.865 *.399 = |-40.644|

Sum of forces (Y) = 0

Sum of forces (X) = Force of woman on gift + Force of friction on gift + Force of gravity (x-component) on gift

Sum of forces (X) = 98.7 +(-40.644) + (-35.274) = 22.782 (x direction)

22.782N = 11kg* a

22.782N/11kg = a = 2.071 m/s^2

Now that we have the acceleration we will be able to use this kinematic equation to solve of the final velocity if we start at vo = 0m/s

v^2 = vo^2 +2a(x)

v^2 = 0 = 2(2.071m/s^2)(8.09m) = 33.50878 m^2/s^2

v = √33.50878m^2/s^2

v = 5.789m/s # final velocity of gift up the slope