How long will it take a 2190 W motor to lift a 1.47 x 104 g box, 6.34 x 104 mm vertically.
1 answer:
Answer:
4.2s
Explanation:
Given parameters:
Power = 2190W
Mass of box = 1.47 x 10⁴g
distance = 6.34 x 10⁴mm
Unknown:
Time = ?
Solution:
Power is the rate at which work is done;
Mathematically;
Power =
Time =
Work done = weight x height
convert mass to kg;
100g = 1kg;
1.47 x 10⁴g = 14.7kg
convert the height to m;
1000mm = 1m
6.34 x 10⁴mm gives 63.4m
Work done = 14.7 x 9.8 x 63.4 = 9133.4J
Time taken = = 4.2s
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Answer:
a) = 928 J
, b)U = -62.7 J
, c) K = 0
, d) Y = 11.0367 m, e) v = 15.23 m / s
Explanation:
To solve this exercise we will use the concepts of mechanical energy.
a) The elastic potential energy is
= ½ k x²
= ½ 2900 0.80²
= 928 J
b) place the origin at the point of the uncompressed spring, the spider's potential energy
U = m h and
U = 8 9.8 (-0.80)
U = -62.7 J
c) Before releasing the spring the spider is still, so its true speed and therefore the kinetic energy also
K = ½ m v²
K = 0
d) write the energy at two points, maximum compression and maximum height
Em₀ = ke = ½ m x²
= mg y
Emo =
½ k x² = m g y
y = ½ k x² / m g
y = ½ 2900 0.8² / (8 9.8)
y = 11.8367 m
As zero was placed for the spring without stretching the height from that reference is
Y = y- 0.80
Y = 11.8367 -0.80
Y = 11.0367 m
Bonus
Energy for maximum compression and uncompressed spring
Emo = ½ k x² = 928 J
= ½ m v²
Emo =
Emo = ½ m v²
v =√ 2Emo / m
v = √ (2 928/8)
v = 15.23 m / s