Answer:
The angle of deflection will be "1.07 × 10⁻⁷°".
Explanation:
The given values are:
Mass of a cow,
m = 1100 kg
Mass of bob,
mb = 1 kg
The total distance between a cow and bob will be,
d = 2 m
Let,
The tension be "t".
The angle with the verticles be "".
Now,
Vertically equating forces
⇒ ...(equation 1)
Horizontally equating forces
⇒ ...(equation 2)
From equation 1 and equation 2, we get
⇒
O putting the estimated values, will be
⇒
⇒
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
Answer:
F= 2.86KN
Explanation:
The attached diagram shows the motion of crate on a single line The forces acting on it have been resolved as shown in the attached figure.The kinetic friction opposes the motion.Now crate is moving with constant speed so the acceleration is constant.
Applying Newton's second law to determine the force F and also substituting the values for to eliminate normal force
F = ma
Now and
components are zero
Horizontal and vertical forces are Force on crate, weight of crate and kinetic friction
therefore
>>> (1)
By putting the above equation in equation 1 we get F
By putting values we get
F =
F= 2860.8 N
F= 2.86KN
