a = 3.6*10^15 m/s^2
accelerating force = M*a = 6.01*10^-12 newtons
Final velocity = V2
Initial Velocity = V1 = 2.4*10^7 m/s
(a) M*V2^2/2 = M*V1^2/2 + F*X
Solve for V2
(b) KE increase = F*X
(a) What is its speed?
(b) What is the increase in its kinetic energy?
accelerating force = M*a = 6.01*10^-12 newtons
Final velocity = V2
Initial Velocity = V1 = 2.4*10^7 m/s
(a) M*V2^2/2 = M*V1^2/2 + F*X
Solve for V2
(b) KE increase = F*X
(a) To find the final speed of the proton, we can use the equation of motion:
v^2 = u^2 + 2as
where:
- v is the final velocity (unknown)
- u is the initial velocity (2.4 x 10^7 m/s)
- a is the acceleration (3.6 x 10^15 m/s^2)
- s is the distance traveled (2.8 cm = 0.028 m)
Substituting the given values into the equation, we can solve for v:
v^2 = (2.4 x 10^7)^2 + 2(3.6 x 10^15)(0.028)
v^2 = 5.76 x 10^14 + 2.016 x 10^14
v^2 = 7.776 x 10^14
To find v, we take the square root of both sides:
v = √(7.776 x 10^14)
v ≈ 8.81 x 10^7 m/s
Therefore, the speed of the proton is approximately 8.81 x 10^7 m/s.
(b) The increase in kinetic energy is given by the formula:
ΔKE = (1/2)mv^2 - (1/2)mu^2
where:
- ΔKE is the increase in kinetic energy (unknown)
- m is the mass of the proton (1.67 x 10^-27 kg)
- v is the final velocity (8.81 x 10^7 m/s)
- u is the initial velocity (2.4 x 10^7 m/s)
Substituting the values into the formula:
ΔKE = (1/2)(1.67 x 10^-27)(8.81 x 10^7)^2 - (1/2)(1.67 x 10^-27)(2.4 x 10^7)^2
ΔKE ≈ 5.24 x 10^-14 Joules
Therefore, the increase in kinetic energy is approximately 5.24 x 10^-14 Joules.
Step 1: Convert the given quantities to SI units:
Mass, m = 1.67 x 10^-27 kg
Acceleration, a = 3.6 x 10^15 m/s^2
Initial speed, u = 2.4 x 10^7 m/s
Distance, s = 2.8 cm = 2.8 x 10^-2 m
Step 2: Find the final speed, using the equation of motion:
v^2 = u^2 + 2as
Plugging in the given values, we have:
v^2 = (2.4 x 10^7 m/s)^2 + 2 * (3.6 x 10^15 m/s^2) * (2.8 x 10^-2 m)
v^2 = 5.76 x 10^14 m^2/s^2 + 2.016 x 10^14 m^2/s^2
v^2 = 7.776 x 10^14 m^2/s^2
Taking the square root of both sides, we get:
v ≈ 8.816 x 10^7 m/s
Therefore, the final speed of the proton is approximately 8.816 x 10^7 m/s.
Step 3: Find the increase in kinetic energy, using the work-energy theorem:
The increase in kinetic energy (ΔKE) of an object is equal to the net work done on it.
The net work done on the proton is given by:
Net work (W) = ΔKE
Using the equation for work, W = F * s, where F is the force applied and s is the displacement, we can find the net work done.
The force applied on the proton is given by:
F = m * a
Plugging in the given values, we have:
F = (1.67 x 10^-27 kg) * (3.6 x 10^15 m/s^2)
F = 6.012 x 10^-12 N
The displacement, s = 2.8 x 10^-2 m
Using the equation W = F * s, we can find the net work done:
W = (6.012 x 10^-12 N) * (2.8 x 10^-2 m)
W = 1.683 x 10^-13 J
Therefore, the increase in kinetic energy, ΔKE, is approximately 1.683 x 10^-13 J.