Suppose air resistance does not exist, roughly what would be the velocity of a 10 lb object 5-seconds into free fall off of a 50
1 answer:
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Complete question is;
Block 1 is resting on the floor with block 2 at rest on top of it. Block 3, at rest on a smooth table with negligible friction, is attached to block 2 by a string that passes over a pulley, as shown in the attachment below. The string and pulley have negligible mass.
Block 1 is removed without disturbing block 2.
Derive an equation for the acceleration of block 3 for any arbitrary values of m3 and m2. Express your answer in terms of m3, m2, and physical constants as appropriate.
Answer:
a = (m2)g/(m3 + m2)
Explanation:
Looking at the attached image, if we consider the free body diagram for block 3, by using Newton's first law of motion, we will arrive at the formula;
T = (m3)a - - - (eq 1)
where;
T is the tension in the string
a is acceleration
m3 is mass of block 3
Meanwhile doing the same with Block 2, the free body diagram would give us the formula; (m2)g - T = (m2)a
Making T the subject gives us;
T = (m2)g - (m2)a - - - (eq 2)
where;
g is acceleration due to gravity
T is the tension in the string
a is acceleration
m2 is mass of block 2
To solve for the acceleration, we will just substitute (m3)a for T in eq 2.
Thus;
(m3)a = (m2)g - (m2)a
(m3)a + (m2)a = (m2)g
a(m3 + m2) = (m2)g
a = (m2)g/(m3 + m2)
Final velocity
Data & Formula:
. final velocity: vf = ?
. initial velocity : vi = 1.9 m/s
. acceleration : a = 0.060m/s^2
. distance : d = 8.5 km
Answer:
g = 0.98 m
Explanation:
given data:
upper portion weight 438 N and its center of gravity is 1.17 m
upper legs weight 144 N and its center of gravity is 0.835 m
lower leg weight 88 N and its center of gravity is 0.270 m
position of center of gravity of whole body can be determine by using following relation
where is mass of respective part and
is center of gravity
g = 0.98 m