(a) Energy is conserved at every point in the block's motion, so the potential energy P stored in the first spring at its maximum compression is the same as is stored in the second spring.
The total work performed on the block by the first spring is
W = -1/2 (110 N/m) (0.21 m²) = -2.4255 J
The work performed by the second spring is the same, so
W = -1/2 (240 N/m) x²
Solve for x :
x² = -2W/(240 N/m) = 0.0202125 m²
x ≈ 0.14 m = 14 cm
(b) By the work-energy theorem, the total work performed by either spring on the block as the spring is compressed is equal to the change in the block's kinetic energy. The restoring force of the spring is the only force involved. At maximum compression, the block has zero velocity, while its kinetic energy and hence speed is maximum just as it comes into contact with either spring.
W = 0 - K
W = -1/2 (0.10 kg) v²
v² = -2W/(0.10 kg) = 48.51 m²/s²
v ≈ 7.0 m/s
R=U/I so
U=RxI
U= 10 x 42
U= 420 volts
Technically friction is acting on the car because it is still rubbing against the street and gravity is pulling the car down preventing it from floating??? lol
Answer:
The work required to pump the liquid out of the spout is 2.6 × 10⁶ J
Explanation:
<u>Step 1:</u> consider a spherical tank, sliced into two equal part, since it is half filled.
The new circular surface has a radius, if we construct a right angled-triangle on the surface. The initial radius of the spherical tank becomes the hypotenuse of the triangle and the radius is calculated as follows;
volume (v)
<u>Step 2:</u> Total height, d, in which the liquid is pumped out;
d = b + 3 + 1
= b + 4
<u>Step 3:</u> Force required to pump the liquid out:
F = mg
But, m = density (ρ) x volume (v)
F = ρvg
Given;
ρ = 900 kg/m³ and g = 9.8 m/s²
<u>Step 4:</u> The work done in pumping the liquid out of the spout
W = F x d
Therefore, the work required to pump the liquid out of the spout is 2.6 × 10⁶ J