Respuesta :
Answer:
The velocity of block = 0.188 [tex]\frac{m}{s}[/tex]
Explanation:
Mass m = 5.6 kg
k = 1040 [tex]\frac{N}{m}[/tex]
[tex]\mu[/tex] = 0.26
[tex]x_{1} =[/tex] 0.035 m , [tex]v_{1}[/tex] = 0
[tex]x_{2} =[/tex] 0.02 m
From work energy theorem
[tex]K_{1} + U_{1} + W_{other} = K_{2} + U_{2}[/tex] --------- (1)
Kinetic energy
[tex]K = \frac{1}{2} k x^{2}[/tex] ------- (1)
Potential energy
[tex]U = \frac{1}{2} k x^{2}[/tex] ------- (2)
Work done
W = F.s ------ (3)
From Newton's second law
[tex]R_{N}[/tex] = mg
[tex]R_{N}[/tex] = 5.6 × 9.81 = 54.9 N
Friction force = 0.4 × 54.9 = 21.9 N
Now the work done by the friction
[tex]W_{f}[/tex] = - 21.9 × 0.015
[tex]W_{f}[/tex] = - 0.329 J
Now kinetic energy
At point 1
[tex]K_{1} = \frac{1}{2} m v^{2} _{1}[/tex]
[tex]K_{1} = 0[/tex]
[tex]U_{1} = \frac{1}{2} k x^{2}[/tex]
[tex]U_{1} = \frac{1}{2} (1040) 0.035^{2}[/tex]
[tex]U_{1} =[/tex] 0.637 J
At point 2
[tex]K_{2} = \frac{1}{2} (5.6) v^{2} _{2}[/tex]
[tex]K_{2} = 2.8 v_{2} ^{2}[/tex]
Potential energy
[tex]U_{2} = \frac{1}{2} k x_2^{2}[/tex]
[tex]U_{2} = \frac{1}{2} (1040) 0.02^{2}[/tex]
[tex]U_{2} = 0.208[/tex] J
From equation (1) we get
0 + 0.637 - 0.329 = 2.8 [tex]v_{2} ^{2}[/tex] + 0.208
2.8 [tex]v_{2} ^{2}[/tex] = 0.1
[tex]v_{2} =[/tex] 0.188 [tex]\frac{m}{s}[/tex]
This is the velocity of block.
