Question

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.

Answer 1

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$V^{2}$

Substitute all the necessary parameters

28 = 1/2 x 5.5 x $V^{2}$

5.5$V^{2}$ = 56

$V^{2}$ = 56/5.5

$V^{2}$ = 10.18

V = $\sqrt{10.1818}$

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

Learn more about work done here: brainly.com/question/8119756