Answer:
Hello your question has some missing parts attached below is the missing part
A object of mass 3.00 kg is subject to a force Fx that varies with position as in the figure below.
answer : speed at x = 5 m = 3.35 m/s
speed at x = 15m = 5.12 m/s
Explanation:
initial speed( x = 0 ) = 2.5 m/s
speed at x = 5.00 m = ?
speed at x = 15 m = ?
Determine speed at x = 5 m
First we will apply the expression for work-energy theorem
w = [tex]\frac{1}{2} m(v^2 - v_{0}^2 )[/tex] ----- ( 1 )
where : w = 7.50J, [tex]v_{0}[/tex] = 2.5m/s , m = 3.0 kg ( in the expression of work )
input values into equation 1
7.50 = 1/2 (3 ) ( v^2 - 2.5^2 )
7.50 = 3/2 ( v^2 - 6.25 )
5 j/kg = v^2 - 6.25
∴ v = √11.25 = 3.35 m/s
Determine speed at X = 15
first we will determine the work done form x = 5 to x = 15
W = 7.5J + 15J + 7.5J = 30J, [tex]v_{0}[/tex] = 2.5m/s , m = 3.0 kg
w = [tex]\frac{1}{2} m(v^2 - v_{0}^2 )[/tex] --- ( 2 )
equation2 becomes
30J = 1/2 ( 3 ) ( v^2 - 6.25 )
30J = 3/2 ( V^2 - 6.25 )
20 J/kg = v^2 - 6.25
v = √26.25 = 5.12 m/s
