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The change in potential energy of the proton is 5.6 x Joule
A uniform electric field is where the electric field strength is the same at all points in the field. In the uniform field, the force experienced by a charge is the same no matter where it is placed in the field.
Given that a proton moves a distance 10 cm in a uniform electric field of 3.5 kN C, in the direction of the field.
- The distance d = 10 cm = 0.1 m
- Electric field E = 3.5 KN/C
- Proton charge q = 1.6 x
C
The Work done = Fd
but F = Eq
Recall that Electric field E = F/q = V/d
Where V = potential difference.
Let us first calculate the V
E = V/d
V = Ed
Substitute all the parameters into the formula above
V = 3.5 × 10³ × 0.1
V = 350 v
from F/q = V/d
make F the subject of formula and substitute it in work formula
F = Vq/d
W.D = Vq/d x d
W.D = Vq
Substitute all the parameters into the formula above
W.D = 350 x 1.6 x
W.D = 5.6 x J
Work done = Energy = Potential Energy
Therefore, the change in potential energy of the proton is 5.6 x <em> Joule</em>
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Answer:
The magnitude of the acceleration of the box is 2.01 m/s².
Explanation:
Hi there!
Please, see the attached figure for a graphical description of the problem.
We have the following horizontal forces:
Fr = friction force.
Fx = Horizontal component of the applied force, F.
And we have the following vertical forces:
Fy = vertical component of the applied force.
N = normal force exerted on the box.
W = weight of the box.
According to Newton´s second law:
∑F = m · a
Then, in the horizontal direction:
Fx - Fr = m · a
Where "m" is the mass of the box and "a" its acceleration.
Fx can be obtained by trigonometry (see figure):
Fx = F · cos 30°
Fx = 90.0 N · cos 30°
Fr is calculated as follows:
Fr = μ · N
Where μ is the coefficient of friction and N the normal force.
So, we have to find the magnitude of the normal force.
Using Newton´s second law in the vertical direction:
∑F = N + Fy - W = m · a
Notice that the box has no vertical acceleration, then:
N + Fy - W = 0
Solving for N:
N = W - Fy
The weight is calculated as follows:
W = m · g
Where g is the acceleration due to gravity:
W = 20.0 kg · 9.8 m/s² = 196 N
And the vertical component of the applied force can be obtained by trigonometry:
Fy = F · sin 30°
Fy = 90.0 N · sin 30°
The normal force will be:
N = W - Fy = 196 N - 90.0 N · sin 30°
N = 151 N
Now, we can calculate the friction force:
Fr = μ · N
Fr = 0.250 · 151 N
Fr = 37.8 N
And now, we can obtain the acceleration of the box:
Fx - Fr = m · a
(Fx - Fr) / m = a
(90.0 N · cos 30° - 37.8 N ) / 20.0 kg = a
a = 2.01 m/s²
The magnitude of the acceleration of the box is 2.01 m/s².
