A horizontal force is applied to a box with a mass equal to 5.5 kg, The graph shows the net force acting on a box as a function of its horizontal position x.

Part A
Find the net work done on the box as it moves from x = 0 m to x = 8.0 m. .
Part B
Apply concepts of work and energy to solve for the speed of the box when it reaches position x = 8.0 m. Assume the box started from rest.

From the graph, net work done on the box and the speed of the box when it reaches position x = 8.0 m are 28 Nm and 3.2 m/s respectively

From the question, these are the given parameters;

Mass M = 5.5 Kg

Position X = 0 to X = 8m

Part A

The net work done will be the area under the graph.

From position X = 2m to X = 8m gives us the shape of a trapezium.

A = 1/2( a + b )h

A = 1/2( 2 + 6 ) x 8

A = 8 x 4

A = 32 Nm

From X = 0 to X = 2 gives us the shape of a triangle.

A = 1/2bh

A = 1/2 x 2 x (-4)

A = -4 x 1

A = -4 Nm

Net Work done = 32 - 4

Net work done = 28 Nm

Part B

Applying the concepts of work and energy to solve for the speed of the box when it reaches position x = 8.0 m

Net Work done = 1/2m[tex]V^{2}[/tex]

Substitute all the necessary parameters

28 = 1/2 x 5.5 x [tex]V^{2}[/tex]

5.5[tex]V^{2}[/tex] = 56

[tex]V^{2}[/tex] = 56/5.5

[tex]V^{2}[/tex] = 10.18

V = [tex]\sqrt{10.1818}[/tex]

V = 3.19 m/s

Therefore, net work done on the box and the speed of the box when it reaches position x = 8.0 m are 28 Nm and 3.2 m/s respectively

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