Answer:
0.102 m
Explanation:
k = spring constant of the spring = 125 N/m
m = mass of the block attached to the spring = 650 g = 0.650 kg
x = maximum extension of the spring
h = height dropped by the block = x
Using conservation of energy
Spring potential energy gained = Gravitational potential energy lost
(0.5) k x² = mgh
(0.5) k x² = mgx
(0.5) (125) x = (0.650) (9.8)
x = 0.102 m
Answer:
According to your question although I think an object undergoing uniform circular motion is moving with a constant speed. Nevertheless, it is accelerating due to its change in direction. The direction of the acceleration is inwards,therefore a force perpendicular to an objects velocity change the direction of the velocity but not its magnitude.
Henry will lift 200 N load 20 m up a ladder in 40 s. While the Ricardo will take 400 N load in 80 seconds. So, For Henry to take 400 N load it will take him 80 seconds in two attempts. And,also, he will have to cover 40 m of distance.
Answer:
Neither.
Explanation:
When an electron is released from rest, in an uniform electric field, it will accelerate moving in a direction opposite to the field (as the field has the direction that it would take a positive test charge, and the electron carries a negative charge).
It will move towards a point with a higher potential, so its kinetic energy will increase, while its potential energy will decrease:
⇒ ΔK + ΔU = 0 ⇒ ΔK = -ΔU = - (-e*ΔV)
As ΔV>0, we conclude that the electric potential energy decreases while the kinetic energy increases in the same proportion, in order to energy be conserved, in absence of non-conservative forces.
Answer: the effective design stiffness required to limit the bumper maximum deflection during impact to 4 cm is 3906250 N/m
Explanation:
Given that;
mass of vehicle m = 1000 kg
for a low speed test; V = 2.5 m/s
bumper maximum deflection = 4 cm = 0.04 m
First we determine the energy of the vehicle just prior to impact;
W_v = 1/2mv²
we substitute
W_v = 1/2 × 1000 × (2.5)²
W_v = 3125 J
now, the the effective design stiffness k will be:
at the impact point, energy of the vehicle converts to elastic potential energy of the bumper;
hence;
W_v = 1/2kx²
we substitute
3125 = 1/2 × k (0.04)²
3125 = 0.0008k
k = 3125 / 0.0008
k = 3906250 N/m
Therefore, the effective design stiffness required to limit the bumper maximum deflection during impact to 4 cm is 3906250 N/m
