Respuesta :
Explanation:
The given data is as follows.
k = 130 N/m, [tex]\Delta x[/tex] = 17 cm = 0.17 m (as 1 m = 100 cm)
mass (m) = 2.8 kg
When the spring is compressed then energy stored in it is as follows.
Energy = [tex]\frac{1}{2}kx^{2}[/tex]
Now, spring energy gets converted into kinetic energy when the box is launched.
So, [tex]\frac{1}{2}kx^{2}[/tex] = [tex]\frac{1}{2}mv^{2}[/tex]
[tex]\frac{1}{2} \times 130 \times (0.17)^{2}[/tex] = [tex]\frac{1}{2} \times 2.8 \times v^{2}[/tex]
[tex]v^{2} = \frac{3.757}{2.8}[/tex]
= 1.34
v = 1.15 m/sec
Now,
Frictional force = [tex]\mu \times mg[/tex]
= [tex]0.15 \times 2.8 \times 9.8[/tex]
= 4.116 N
Also, Kinetic energy = work done by friction
[tex]\frac{1}{2}mv^{2} = F_{f} \times d[/tex]
[tex]\frac{1}{2} \times 2.8 \times (1.15)^{2} = 4.116 \times d[/tex]
1.8515 = [tex]4.116 \times d[/tex]
d = 0.449 m
Thus, we can conclude that the box slides 0.449 m across the rough surface before stopping.
