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.
The answer to this question is A - 25 N
Answer:
W = 0.060 J
v_2 = 0.18 m/s
Explanation:
solution:
for the spring:
W = 1/2*k*x_1^2 - 1/2*k*x_2^2
x_1 = -0.025 m and x_2 = 0
W = 1/2*k*x_1^2 = 1/2*(250 N/m)(-0.028m)^2
W = 0.060 J
the work-energy theorem,
W_tot = K_2 - K_1 = ΔK
with K = 1/2*m*v^2
v_2 = √2*W/m
v_2 = 0.18 m/s
Answer:
Explanation:
From the given information:
The initial PE 
= m×g×h
= 5 kg × 9.81 m/s² × 10 m
= 490.5 J
The change in Potential energy P.E of the box is:
ΔP.E = 
ΔP.E = 0 -
ΔP.E = 
If we take a look at conservation of total energy for determining the change in the internal energy of the box;
this can be re-written as:
Here, K.E = 0
Also, 70% goes into raising the internal energy for the box;
Thus,
ΔU = 343.35 J
Thus, the magnitude of the increase is = 343.35 J
