The electric force acting on the charge is given by the charge multiplied by the electric field intensity:
where in our problem
and
, so the force is
The initial kinetic energy of the particle is zero (because it is at rest), so its final kinetic energy corresponds to the work done by the electric force for a distance of x=4 m:
Explanation:
Given that,
Distance, s = 47 m
Time taken, t = 8.6 s
Final speed of the truck, v = 2.3 m/s
Let u is the initial speed of the truck and a is its acceleration such that :
.............(1)
Now, the second equation of motion is :
Put the value of a in above equation as :
u = 8.63 m/s
So, the original speed of the truck is 8.63 m/s. Hence, this is the required solution.
Answer:
The kinetic energy of the proton at the end of the motion is 1.425 x 10⁻¹⁶ J.
Explanation:
Given;
initial velocity of proton, 
= 3 x 10⁵ m/s
distance moved by the proton, d = 3.5 m
electric field strength, E = 120 N/C
The kinetic energy of the proton at the end of the motion is calculated as follows.
Consider work-energy theorem;
W = ΔK.E
where;
K.Ef is the final kinetic energy
W is work done in moving the proton = F x d = (EQ) x d = EQd
Therefore, the kinetic energy of the proton at the end of the motion is 1.425 x 10⁻¹⁶ J.
