Respuesta :
Answer:
a) T = 2838.6 N, b) W = 1003.2 J, c) W = 6.22 10⁴ J, d) W = 2.79 10³ J
e) v_f = 2.65 m / s
Explanation:
a) To find the tension of the cable let's use Newton's second law
T - W = m a
T = W + ma
T = m (g + a)
let's calculate
T = 285 (-9.8 - 0.160)
T = 2838.6 N
b) net work is stress work minus weight work
W = F d
W = (T-W) d
W = (m a) d
W = (285 0.160) 22
W = 1003.2 J
c) the work done by the cable
W = T d cos 0
W = 2838.6 22.0
W = 6.22 10⁴ J
d) The work done by the weight
the displacement is upwards and the weight points downwards, so the angle is 180º
W = F. d
W = F d cos 180
W = -285 22.0
W = 2.79 10³ J
e) the final speed of the load. Let's use the relationship between work and the change in kinetic energy
W = ΔK
as part of rest K₀ = 0
W = ½ m v_f²
v_f = [tex]\sqrt{ \frac{2W}{m} }[/tex]
v_f = [tex]\sqrt{\frac{2 \ 1003.2}{285} }[/tex]
v_f = 2.65 m / s
