Answer:
a) The maximum distance compressed by the spring = 0.6m
b) 1.5m/s
Explanation:
mass of the block m= 5.00kg
Initial velociy of the block (v₀) = 6.00m/s
spring constant of the block is (k) = 500N/m
a) The maximum compression in the spring can be calculated as
initial kinetic energy is given by
[tex]K_i = \frac{1}{2} mv^2_0[/tex] -----------------(1)
Final kinetic energy of the block is given by
[tex]K_f = \frac{1}{2} mv^2_f[/tex] ------------------(2)
Change in kinetic energy of the block
ΔKE = [tex]K_f - K_i[/tex]
= [tex]\frac{1}{2} (5.00)(0)^2 - \frac{1}{2} (5.00) (6.00)^2\\\\= -90J[/tex]
The work done by the spring is stored in the form of potential energy of the block
[tex]U_s = \frac{1}{2} kx_m_a_x^2 ------------- (3)[/tex]
[tex]x_m_a_x[/tex] is the maximum compressed distance of the spring
The loss in the kinectic energy will be converted to the potential energy of the block
ΔKE = [tex]U_s[/tex]
[tex]90J = \frac{1}{2} kx_m_a_x^2[/tex]
[tex]x_m_a_x = \sqrt{\frac{2(90)}{500} } \\\\= 0.6m[/tex]
The maximum distance compressed by the spring = 0.6m
b) The maximum compressed in the spring = 0.150m
let the initial velocity of the block be v
[tex]\frac{1}{2} mv^2 = \frac{1}{2} kx_m_a_x^2[/tex]
therefore, the initial velocity of the block
[tex]v = \sqrt{\frac{kx_m_a_x^2}{m} } \\\\v = \sqrt{\frac{500 \times 0.150 }{5.00} } \\\\= 1.5m/s[/tex]
(a) The maximum distance compressed by the spring 0.6m.
(b) For the spring is to compress by no more than 0.150 m, then the maximum value of speed is 1.5 m/s.
Given data:
The mass of block is, m = 5.00 kg.
The speed of block is, v = 6.00 m/s.
The spring force constant is, k = 500 N/m.
(a)
When compression of spring takes place, the kinetic energy of spring is stored in the form of spring potential energy. Then the relation is,
Kinetic energy = Spring potential energy
[tex]\dfrac{1}{2}m \times v^{2} = \dfrac{1}{2}k \times x^{2}[/tex]
Here, x is the compressed distance.
Solving as,
[tex]5 \times 6^{2} = 500 \times x^{2}\\\\x = \sqrt{\dfrac{5 \times 36}{500}}\\\\x = 0.6 \;\rm m[/tex]
Thus, the maximum distance compressed by the spring 0.6m.
(b)
In this part, we need to calculate the speed with compressed distance of 0.150 m. Then,
Kinetic energy = Spring potential energy
[tex]\dfrac{1}{2}m \times v^{2} = \dfrac{1}{2}k \times x^{2}\\\\m \times v^{2} =k \times x^{2}\\\\5.00 \times v^{2} =500 \times 0.150^{2}\\\\v=\sqrt{\dfrac{500 \times 0.150^{2}}{5.00}}\\\\v=1.5 \;\rm m/s[/tex]
Thus, If the spring is to compress by no more than 0.150 m, then the maximum value of speed is 1.5 m/s.
Learn more about the spring constant here:
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