A net force of 1.6×10−15 N acts on an electron over a displacement of 5.0 cm, in the same direction as the net force. (a) What is the change in kinetic energy of the electron in joules? (b) If the electron was initially at rest, what is the speed of the electron in m/s? An electron has a mass of 9.1×10−31 kg.
To solve this problem, we need to use the equations of motion and the concept of work and energy.
(a) What is the change in kinetic energy of the electron in joules?
The work done on an object is given by the formula:
Work = Force × Displacement × cos(θ)
In this case, since the force and the displacement are in the same direction, the angle θ = 0°, and cos(0°) = 1.
Therefore, the work done on the electron is:
Work = (1.6×10^-15 N) × (5.0 cm) = (1.6×10^-15 N) × (5.0 × 10^-2 m) = 8.0 × 10^-17 J
Now, the work done on an object is equal to the change in kinetic energy of that object. Therefore, the change in kinetic energy of the electron is 8.0 × 10^-17 J.
(b) If the electron was initially at rest, what is the speed of the electron in m/s?
The kinetic energy of an object is given by the formula:
Kinetic Energy = 1/2 × mass × velocity^2
Since the electron was initially at rest, its initial kinetic energy is zero:
Initial Kinetic Energy = 0
The final kinetic energy of the electron is equal to the change in kinetic energy calculated in part (a):
Final Kinetic Energy = 8.0 × 10^-17 J
Setting the initial and final kinetic energy equal, we have:
0 = 1/2 × (9.1×10^-31 kg) × velocity^2
Solving for the velocity, we have:
velocity^2 = (2 × Final Kinetic Energy) / mass = (2 × 8.0 × 10^-17 J) / (9.1×10^-31 kg)
velocity^2 ≈ 3.518 × 10^13 m^2/s^2
Taking the square root of both sides, we find:
velocity ≈ 1.875 × 10^6 m/s
Therefore, the speed of the electron is approximately 1.875 × 10^6 m/s.