The potential energy of the spring is 6.75 J
The elastic potential energy stored in the spring is given by the equation:
where;
k is the spring constant
x is the compression/stretching of the string
In this problem, we have the spring as follows:
k = 150 N/m is the spring constant
x = 0.3 m is the compression
Substituting in the equation, we get
Therefore. the elastic potential energy stored in the spring is 6.75J .
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Answer:
(a) Acceleration of positron is 6.03 x 10¹³ m/s²
(b) Speed of positron after 8.70 x 10⁻⁹ s is 5.24 x 10⁵ m/s
Explanation:
Given :
Constant electric field, E = 343 N/C
Mass of positron, m = 9.1 x 10⁻³¹ kg
Charge of positron, q = +e = 1.6 x 10⁻¹⁹ C
(a) Coulomb force on the positron is determine by the relation :
F = q x E ....(1)
But, force is also equals to product of mass and acceleration. So,
F = ma .....(2)
Here a is acceleration.
From equation (1) and (2).
m x a = q x E
Substitute the values of q, E and m in the above equation.
a = 6.03 x 10¹³ m/s²
(b) Initially, the positron is at rest, so its initial speed is zero.
The equation of motion for positron is :
v = u + at
Here v is final speed, u is initial speed and t is time.
Since, u is zero, so the equation becomes :
v = at
Substitute 8.70 x 10⁻⁹ s for t and 6.03 x 10¹³ m/s² for a in the above equation.
v = 6.03 x 10¹³ x 8.70 x 10⁻⁹ m/s
v = 5.24 x 10⁵ m/s