How much force was applied to an object if was moved 2 meters and the work done on the object was 8 joules?
2 answers:
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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).
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).
Answer:
0.087 m
Explanation:
Length of the rod, L = 1.5 m
Let the mass of the rod is m and d is the distance between the pivot point and the centre of mass.
time period, T = 3 s
the formula for the time period of the pendulum is given by
.... (1)
where, I is the moment of inertia of the rod about the pivot point and g is the acceleration due to gravity.
Moment of inertia of the rod about the centre of mass, Ic = mL²/12
By using the parallel axis theorem, the moment of inertia of the rod about the pivot is
I = Ic + md²
Substituting the values in equation (1)
12d² -26.84 d + 2.25 = 0
d = 2.15 m , 0.087 m
d cannot be more than L/2, so the value of d is 0.087 m.
Thus, the distance between the pivot and the centre of mass of the rod is 0.087 m.