A forrestry crane is lifting vertically a 2 tons log with a

force F = ( 9 + 2 h) where h is the height.
Draw the
free body diagram with the forces and use work and
energy conservation to determine the speed of the log
when it has risen h=6m.
Apply kinematics to give the expression of the integral
for the time it takes to reach that height starting from
rest?

To solve this problem, let's start by drawing the free body diagram of the log being lifted by the forestry crane:

1. Draw a dot to represent the center of mass of the log.
2. Label the dot as "log" and draw an arrow pointing upwards from the dot to represent the force of the crane, F.
3. Since the log is in equilibrium, draw an arrow pointing downwards from the dot to represent the weight of the log, mg, where m is the mass of the log and g is the acceleration due to gravity.
4. Label the downward arrow as "mg".

Now, let's use work and energy conservation to determine the speed of the log when it has risen to a height of h = 6m.

We know that the work done on an object is equal to the change in its kinetic energy. In this case, the net work done on the log is given by the force F multiplied by the distance h.

Work = Force x Distance
Work = Fh

The change in kinetic energy of the log is given by:

Change in Kinetic Energy = (1/2)mv^2 - (1/2)m(0)^2
Change in Kinetic Energy = (1/2)mv^2 - 0
Change in Kinetic Energy = (1/2)mv^2

Since the work done on the log is equal to the change in its kinetic energy, we can write:

Fh = (1/2)mv^2

Substituting the given force expression, F = 9 + 2h, we have:

(9 + 2h)h = (1/2)mv^2

Now, let's substitute the given mass of the log which is 2 tons or 2000 kg:

(9 + 2h)h = (1/2)(2000)v^2

Simplifying the equation, we have:

9h + 2h^2 = 1000v^2

Since we are interested in finding the speed of the log when it reaches a height of h = 6m, we can substitute h = 6 and solve for v:

9(6) + 2(6^2) = 1000v^2
54 + 72 = 1000v^2
126 = 1000v^2
v^2 = 0.126
v ≈ 0.355 m/s

Therefore, the speed of the log when it has risen to a height of h = 6m is approximately 0.355 m/s.

Now, let's move on to the second question and apply kinematics to give the expression of the integral for the time it takes to reach that height starting from rest.

We can use the equation of motion for an object starting from rest:

h = (1/2)gt^2

In this equation, h represents the height, g represents the acceleration due to gravity, and t represents the time.

Since we are given h = 6m, we can rewrite the equation as:

6 = (1/2)(9.8)t^2
12 = 9.8t^2

To solve for t, we can take the square root of both sides:

t^2 = 12/9.8
t ≈ √(12/9.8)
t ≈ √(1.224)

Therefore, the expression for the integral to determine the time it takes to reach a height of h = 6m starting from rest is approximately √(1.224).