Answer:
(a) [tex]W_c=127.008 J[/tex]
(b) [tex]W_g=148.176 J[/tex]
(c) K.E. = 21.168 J
(d) [tex]v=3.4293m.s^{-1}[/tex]
Explanation:
Given:
[tex]\Rightarrow[/tex] That a constant upward acceleration of [tex]\frac{6g}{7}[/tex] is applied in the presence of gravity.
∴[tex]a=- \frac{6g}{7}[/tex]
(a)
Force by the cord on the block,
[tex]F_c= M\times a[/tex]
[tex]F_c=3.6\times (-6)\times\frac{9.8}{7}[/tex]
[tex]F_c=-30.24 N[/tex]
∴Work by the cord on the block,
[tex]W_c= F_c\times d[/tex]
[tex]W_c= -30.24\times 4.2[/tex]
We take -ve sign because the direction of force and the displacement are opposite to each other.
[tex]W_c=-127.008 J[/tex]
(b)
Force on the block due to gravity:
[tex]F_g= M.g[/tex]
∵the gravity is naturally a constant and we cannot change it
[tex]F_g=3.6\times 9.8[/tex]
[tex]F_g=35.28 N[/tex]
∴Work by the gravity on the block,
[tex]W_g=F_g\times d[/tex]
[tex]W_g=35.28\times 4.2[/tex]
[tex]W_g=148.176 J[/tex]
(c)
Kinetic energy of the block will be equal to the net work done i.e. sum of the two works.
mathematically:
[tex]K.E.= W_g+W_c[/tex]
[tex]K.E.=148.176-127.008[/tex]
K.E. = 21.168 J
(d)
From the equation of motion:
[tex]v^2=u^2+2a_d\times d[/tex]
putting the respective values:
[tex]v=\sqrt{0^2+2\times \frac{9.8}{7}\times 4.2 }[/tex]
[tex]v=3.4293m.s^{-1}[/tex] is the speed when the block has fallen 4.2 meters.