Respuesta :
Answer:
V=15.3 m/s
Explanation:
To solve this problem, we have to use the energy conservation theorem:
[tex]U_e+K_i+U_{gi}+W_{friction}=K_f+U_{gf}[/tex]
the elastic potencial energy is given by:
[tex]U_e=\frac{1}{2}*k*x^2\\U_e=\frac{1}{2}*1100N/m*(4m)^2\\U_e=8800J[/tex]
The work is defined as:
[tex]W_{friction}=F_f*d*cos(\theta)\\W_{friction}=40N*2.5m*cos(180)\\W_{friction}=-100J[/tex]
this work is negative because is opposite to the movement.
The gravitational potencial energy at 2.5 m aboves is given by:
[tex]U_{gf}=m*g*h\\U_{gf}=60kg*9.8*2.5\\U_{gf}=1470J[/tex]
the gravitational potential energy at the ground and the kinetic energy at the begining are 0.
[tex]8800J+0+0+-100J=\frac{1}{2}*62kg*v^2+1470J\\v=\sqrt{\frac{2(8800J-100J-1470J)}{62kg}}\\v=15.3m/s[/tex]
The speed that the Great Sandini will emerge from the end of the barrel, 2.5m above his initial rest position is 15.3 m/s.
What is the energy conservation theorem?
According to the law of conservation of energy, energy can neither be created nor destroyed - only converted from one form of energy to another.
In order to solve the problem we have to use the energy conservation theorem, therefore,
[tex]\rm U_e+K_i + U_{gi}+W_{friction} = K_f + U_{gf}[/tex]
The elastic potential energy can be given by the formula,
[tex]U_e = \frac{1}{2} \times k \times x^2[/tex]
Substitute the value of the spring constant k= 1100 N/m, while the displacement is 4m.
[tex]U_e = \dfrac{1}{2} \times 1100 \times 4^2\\\\U_e = 8800\rm\ J[/tex]
The work done can be defined as,
[tex]W_{friction} =\rm F_f\cdot d \cdot\ cos\theta[/tex]
Substitute the value of the friction force as 40, displacement(d) as 2.5 while the measure of the angle is 180°.
[tex]W_{friction} =\rm 40\times 2.5 \times\ cos(180^o)\\\\W_{friction} = -100\ J[/tex]
We got the work is negative because the friction force is applied in the opposite direction to the movement.
The gravitational potential energy at the height of 2.5 m above can be given as,
[tex]U_{gf} = mgh\\\\U_{gf} = 60 \times 9.81 \times 2.5\\\\U_{gf} = 1470\rm\ J[/tex]
Substituting all the values in the formula of the energy conservation theorem, also, gravitational potential energy at the ground and the kinetic energy at the beginning as 0. therefore,
[tex]\rm U_e+K_i + U_{gi}+W_{friction} = K_f + U_{gf}\\\\\rm U_e+K_i + U_{gi}+W_{friction} = (\frac{1}{2}mv^2)+ U_{gf}\\\\\rm 8800+0+ 0-100= (\dfrac{1}{2}\times 62 \times v^2 )+ 1470\\\\v = 15.3\ m/s[/tex]
Hence, the speed that the Great Sandini will emerge from the end of the barrel, 2.5m above his initial rest position is 15.3 m/s.
Learn more about the Energy conservation theorem:
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