Respuesta :
Answer:
a) v = 11.3 m/s
b) d = 8.83 m
c) v = 11.3 m/s and d = 8.83 m
Explanation:
a) The work-energy theorem states that the work done on an object is equal to the change in the kinetic energy of the object.
[tex]W = K_2 - K_1\\Fx = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2\\-mg\mu_k x = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2\\-5*9.8*0.27*3 = \frac{1}{2}5v_2^2 - \frac{1}{2}5(12)^2\\v_2 = 11.3~m/s[/tex]
b) We will use the conservation of energy.
[tex]K_2 + U_2 - W_{friction} = K_3 + U_3\\\frac{1}{2}mv_2^2 + 0 - mg\cos(35^\circ)\mu d = 0 + mgh\\h = d\sin(35^\circ)\\\frac{1}{2}5(11.3)^2 - 5(9.8)\cos(35)0.2d = 5(9.8)d\sin(35)\\319.2 - 8.02d = 28.1d\\319.2 = 36.12d\\d = 8.83 m[/tex]
c) Newton’s Second Law will be applied.
[tex]F_{net} = ma \\-mg\mu = ma \\a = -g\mu = -2.64[/tex]
The final speed can be found by using the kinematics equations:
[tex]v_2^2 = v_1^2 + 2ax\\v_2^2 = (12)^2 - 2(2.64)3//\\v_2 = 11.3~m/s[/tex]
Similarly,
[tex]F_{net} = ma\\-mg\cos(35)\mu - mg\sin(35) = ma \\-g\cos(35)\mu - g\sin(35) = a\\-9.8\cos(35)0.2 - 9.8\sin(35) = a\\a =-7.22~m/s^2[/tex]
The kinematics equations are
[tex]v_3^2 = v_2^2 + 2ad\\0 = (11.3)^2 - 2(7.22)d\\d = 8.83m[/tex]
