What is the work done by tension before the block goes up the incline?

1) What is the work done by tension before the block goes up the incline? (On the horizontal surface.)

Work = 75 * 7.7 = 577.5

2) What is the speed of the block right before it begins to travel up the incline?

The work causes the block’s kinetic energy to increase. Since the block’s initial velocity is 0 m/s, the block’s final kinetic energy is equal to the work.

½ * 16 * vf^2 = 577.5

vf = √577.5/8 = √72.1875

This is approximately 8.5 m/s.

Let’s determine the forces which affect the object’s acceleration. Since the object is moving up the incline, the friction force and the component of the object’s that is parallel to the incline are causing the object to decelerate. Let’s determine the magnitude of these two forces.

Ff = 0.43 * 16 * 9.8 * cos 34 = 67.424 * cos 34

Force parallel = 16 * 9.8 * sin 34 = 156.8 * sin 34

3) What is the work done by friction after the block has traveled a distance x = 3.6 m up the incline.

Work = 67.424 * cos 34 * 3.6 = 242.7264 * cos 34

This is approximately 201.2 N *m

4) What is the work done by gravity after the block has traveled a distance x = 3.6 m up the incline? (Where x is measured along the incline.)

The done by gravity is equal to the increase of the block’s potential energy. To determine this we need to determine the increase of the block’s height. Use the following equation.

∆h = 3.6 * sin 34

∆PE = 16 * 9.8 * 3.6 * sin 34 = 564.48 * sin 34

This is approximately 315.65 J

5) How far up the incline does the block travel before coming to rest? (Measured along the incline)

To determine the distance the block travel before coming to rest, use the following equation.

vf^2 = vi^2 + 2 * a * d

vf = 0

vi is the block’s velocity right before it begins to travel up the incline.

vi^2 = 72.1875

0 = 72.1875 + 2 * a * d

Let’s determine the forces which affect the object’s acceleration. Since the object is moving up the incline, the friction force and the component of the object’s that is parallel to the incline are causing the object to decelerate. Let’s determine the magnitude of these two forces.

Ff = 0.43 * 16 * 9.8 * cos 34 = 67.424 * cos 34

Force parallel = 16 * 9.8 * sin 34 = 156.8 * sin 34

Net force = mass * acceleration

Net force = 75 – 67.424 * cos 34 – 156.8 * sin 34 = 16 * a

a = (75 – 67.424 * cos 34 – 156.8 * sin 34) ÷ 16 = -4.286154785 m/s^2

Let’s use this number in the equation and solve for d.

0 = 72.1875 + 2 * -4.286154785 * d

-72.1875 = -8.57230957 * d

d = -72.1875 ÷ -8.57230957

This is approximately 8.421 meters.

1) Work = force x distance

2) Work = change in energy

Assuming the block starts at rest, what is its kinetic energy right before the incline?

3) To find the work you first need to find the force of friction, for which you need the magnitude of the Normal force.

With the x-axis tilted up the incline, the y-axis is in the direction of the Normal force. Note that only gravity has both x and y components. Calculate them.

The y-component of the net force is zero. What must the Normal force be?

4) Work = force x distance. To multiply vectors you need to take into account their direction. What is the distance traveled in the direction of gravity?

Equivalently, Work = change in energy. What is the change in gravitational potential energy?

5) When the block travels a distance x up the incline, how much work has been done by:

friction?

gravity?

tension?

normal force?

Note that some of these are negative. What is the block’s energy (starting energy + work done)?

At what x does the energy equal 0?