Physics Graph question -- Pushing a block on a frictionless surface

[Bdb1331]

The diagram shows how a force along the x-axis varies with location. A block of mass 7.20 kg moves along the x-axis on a horizontal frictionless surface, influenced by this force. (Graph is attached)
A) How much work is done by the force as the block moves from the origin to 8.0 m?

B) If the block has a velocity of 3.80 m/s at the origin, what is its velocity at 8.0 m?

For part A, I worked out the work to be 33 J by finding the area under the curve...
in the first two seconds

W = F(avg)d
W = [(2 + 8) / 2](2 - 0)
W = 10 J

from 2 to 4 seconds

W = [(8 + 4)/2] (4 - 2)
W = 12 J

from 4 to 6 s

W = 4(6 - 4)
W = 8 J

from 6 to 8 s

W = [(4 + -1)/2](8 - 6)
W = 3 J

Total work

W = 10 + 12 + 8 + 3
W = 33 J

But I am sort of lost when it comes to B)

 

[PhanthomJay]

Are you familiar with work-energy relationships?

 

[Bdb1331]

Are you familiar with work-energy relationships?

Sorta, but not exactly confident in my skills

 

[PhanthomJay]

Sorta, but not exactly confident in my skills

Well, just try using the work - energy theorem. What does that theorem say?

 

[HalfLostAndSearching]

For part B, we can use the work-energy theorem to solve for the final velocity of the block. The work-energy theorem states that the work done by a net force on an object is equal to the change in kinetic energy of that object. In this case, the net force is the force shown in the graph, and the change in kinetic energy is the difference between the initial and final kinetic energies.

Initial kinetic energy = 1/2 * m * v^2 = 1/2 * 7.20 kg * (3.80 m/s)^2 = 51.84 J

Final kinetic energy = 1/2 * m * v^2 = 1/2 * 7.20 kg * v^2

Using the work-energy theorem, we can set these two expressions equal to each other and solve for v:

Work done by force = Change in kinetic energy
33 J = 1/2 * 7.20 kg * v^2 - 51.84 J
84.84 J = 1/2 * 7.20 kg * v^2
v^2 = 84.84 J / (1/2 * 7.20 kg)
v^2 = 23.567 m^2/s^2
v = √23.567 m/s
v = 4.855 m/s

Therefore, the velocity of the block at 8.0 m is 4.855 m/s.