A force diagram of the given situation is shown below:
As you can notice, tension force T and weight m*g of the object are opposite.
If the object goes downward with an acceleration 2m/s^2, it means that there is a net force, given by:
Fnet = m*g - T = m*a
m: mass of the obejct = 215 kg
a: acceleration = 2 m/s^2
g: gravitational acceleration constant = 9.8m/s^2
(a) Solve the equation above for T and replace the values of the other parameters:
T = m*g - m*a
T = (215kg)(9.8m/s^2) - (215kg)(2m/s^2)
T = 1677N
(b) The net work is given by:
Wnet = Wweight - Wtension
Wtension is the work done by the tension and Wweight is the work done by the weight of the object.
Wtension = T*y
Wweight = m*g*y
y is the vertical distance traveled by the object, which is y = 20m. Replace the values of m, g, y and T into the previous expressions:
Wtension = (1677N)(20m) = 33540J
Wweight = (215kg)(9.8m/s^2)(20m) = 42140J
Wnet = 42140J - 33540J = 8600J
(c) The work done by the cable is the work done by the tension:
Wcable = Wtension = 33540J
(d) Work done by gravity is the work done by the weight of the object:
Wgravity = Wweight = 42140J
(e) Use the following kinematic formula for the final speed of the obejct:
[tex]v^2=v^2_o+2gy[/tex]
where,
vo: initial speed = 0m/s
v: final speed = ?
y: vertical distance = 20m
g: gravitational acceleration constant = 9.8m/s^2
Solve for v and replace the values of the other parameters and simplify:
[tex]\begin{gathered} v=\sqrt[]{2gy} \\ v=\sqrt[]{2(9.8\frac{m}{s^2})(20m)} \\ v\approx19.80\frac{m}{s} \end{gathered}[/tex]
