Answer:
true is the answer of the question
Refer to the figure shown below.
Let m₁ and m₂ e the two masses.
Let a = the acceleration.
Let T = tension over the frictionless pulley.
Write the equations of motion.
m₂g - T = m₂a (1)
T - m₁g = m₁a (2)
Add equations (1) and (2).
m₂g - T + T - m₁g = (m₁ + m₂)a
(m₂ - m₁)g = (m₁ + m₂)a
Divide through by m₁.
(m₂/m₁ - 1)g = (1 + m₂/m₁)a
Define r = m₂/m₁ as the ratio of the two masses. Then
(r - 1)g = (1 +r)a
r(g-a) = a + g
r = (g - a)/(g + a)
With = 2 ft/s from rest, the acceleration is
a = 2/32.2 = 0.062 ft/s²
Therefore
r = (32.2 - 0.062)/(32.2 + 0.062) = 0.9962
Answer:
The ratio of masses is 0.9962 (heavier mass divided by the lighter mass).
It makes no sense how you typed this problem out.
Answer:
25032.47 W
Explanation:
Power is the time rate of doing work, hence,
P = Work done(non conservative) / time
Work done (non conservative) is given as:
W = total K. E. + total P. E.
Total K. E. = 0.5mv²- 0.5mu²
Where v (final velocity) = 7.0m/s, u (initial velocity) = 0m/s
Total P. E. = mgh(f) - mgh(i)
Where h(f) (final height) = 7.2m, h(i) (initial height) = 0 m
=> W = 0.5mv² - mgh(f)
P = [0.5mv² - mgh(f)] / t
P = [(0.5*790*7²) - (790*9.8*7.2)] / 3
P = (19355 + 55742.4) / 3 = 75097.4/3
P = 25032.47 W