Answer:
(a) -364.64 J
(b) +410.2 J
(c) 45.58 J
(d) 3.17 m/s
Explanation:
We apply Newton's second law:
∑F = m*a (Formula 1)
∑F : algebraic sum of the forces in Newton (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
Data:
m= 9.1 kg
ay= g/9 m/s²
g = 9.81m/s² : acceleration due to gravity
W= m*g : Block weight
W= m*g =9.1 kg*9.8 m/s²= 89.18 N
d= 4.6 m
Problem development
We apply the formula 1 to calculate the cord's force on the block,(T):
∑Fy = m*ay
W-T = m*g/9
W-T = W/9
T = W-W/9=(8/9) W= (8/9) *89.18 = 79.27 N
(a) Work done by the cord's force on the block (Wc)
Wc= - T*d = - 79.27 N*4.6 m = - 364.64 J
The work ,Wc, is negative because the force (T) goes in the opposite direction to the block displacement.
(b)Work done by the gravitational force on the block(Wb)
Wb = W*d= 89.18 N*4.6 m = +410.2 J
The work ,Wb , is positive because the force (W) goes in the same direction as the block displacement
(c) kinetic energy of the block (K)
K =( 1/2)*m*v²
v² = v₀²+2*a*d ,v₀=0 , v²=2*a*d= 2*(g/9)* 4.6
K= ( 1/2)* 9.1* 2*(9.8/9)* 4.6 = 9.1* (9.8/9)* 4.6= 45.58 J
(d) the speed of the block.(v)
v²=2*a*d= 2*(g/9)* 4.6= 2*(9.8/9)* 4.6
[tex]v= \sqrt{\frac{2*9.8*4.6}{9} }[/tex]
v= 3.17 m/s
