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
Substitute all the necessary parameters
28 = 1/2 x 5.5 x
5.5 = 56
= 56/5.5
= 10.18
V =
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