Respuesta :
Answer
given,
force constant = 130 N/m
mass of block = 4 kg
μk = 0.400
magnitude of force = 82 N
compressed to the distance of (x) = 80 cm = 0.8 m
using conservation of energy
[tex]F x = \dfrac{1}{2}kx^2+\dfrac{1}{2}mv^2+\mu_kmgx[/tex]
[tex]\dfrac{1}{2}mv^2 = F x - \dfrac{1}{2}kx^2-\mu_kmgx[/tex]
[tex]mv^2 = 2Fx-kx^2-2\mu_kmgx[/tex]
[tex]v=\sqrt{\dfrac{2Fx}{m}-\dfrac{kx^2}{m}-2\mu_kmgx}[/tex]
[tex]v=\sqrt{\dfrac{2\times 82\times 0.8}{4}-\dfrac{130\times 0.8^2}{4}-2\times 0.4\times 9.8\times 0.8}[/tex]
v=2.39 m/s
Net force on the block is
[tex]F_{net} = F - F_f -kx[/tex]
= 82 - 0.4×4×9.8 - 130×0.8
= -37.68 N
Acceleration of the block
[tex]a = \dfrac{F_{net}}{m}[/tex]
[tex]a = \dfrac{37.68}{4}[/tex]
a =9.42 m/s²
The speed of block at the instant the spring was compressed is 2.39 m/s.
The magnitude of the block's acceleration is 3.57 m/s² and it is opposite to the direction of the spring compression.
The given parameters;
- mass of the block, m = 4 kg
- force constant, k = 130 N/m
- coefficient of kinetic friction, μk = 0.4
- the force applied on, F = 82 N
- compression of the spring, x = 80 cm = 0.8 m
Apply the principle of work-energy theorem, to determine the speed of block when it was compressed.
[tex]work-done = Energy\\\\Fx = \frac{1}{2} mv^2 + \frac{1}{2} kx^2 + F_k \mu_k x \\\\\frac{1}{2} mv^2 = Fx - \frac{1}{2} kx^2 - F_k \mu_k x\\\\mv^2 = 2Fx - kx^2 - 2F_k\mu_kx\\\\v^2 = \frac{2Fx - kx^2 - 2F_k\mu_kx}{m} \\\\v = \sqrt{\frac{2Fx - kx^2 - 2F_k\mu_kx}{m} } \\\\v = \sqrt{\frac{2Fx - kx^2 - 2mg\mu_kx}{m} }\\\\v = \sqrt{\frac{2(82\times 0.8) - (130\times 0.8^2) - 2(4\times 9.8 \times 0.4 0.8)}{4} }\\\\v =2.39 \ m/s[/tex]
The acceleration of the block is calculated as;
[tex]v^2 = u^2 + 2a(x)\\\\ 0 = u^2 + 2ax\\\\a = \frac{u^2}{-2x} \\\\a = \frac{(2.39)^2}{-2(0.8)} \\\\a = -3.57 \ m/s^2[/tex]
Thus, the magnitude of the block's acceleration is 3.57 m/s² and it is opposite to the direction of the compression.
Learn more here:https://brainly.com/question/10063455
