For this case, the first thing you should do is write the kinematic motion equation of the block.
We have then:
vf = vo + a * t
Where,
vf: Final speed.
vo: Initial speed.
a: acceleration.
t: time.
Substituting the values:
(16) = (0) + a * (16)
Clearing the acceleration:
a = 16/16 = 1m / s ^ 2
Note: the other data for this case are not used in this problem.
answer:
The acceleration of the box is 1m / s ^ 2
The magnetic force experienced by the proton is given by

where q is the proton charge, v its velocity, B the magnitude of the magnetic field and

the angle between the direction of v and B. Since the proton moves perpendicularly to the magnetic field, this angle is 90 degrees, so

and we can ignore it in the formula.
For Netwon's second law, the force is also equal to the proton mass times its acceleration:

So we have

from which we can find the magnitude of the field:
Answer:
Explanation:
Given that,
5J work is done by stretching a spring
e = 19cm = 0.19m
Assuming the spring is ideal, then we can apply Hooke's law
F = kx
To calculate k, we can apply the Workdone by a spring formula
W=∫F.dx
Since F=kx
W = ∫kx dx from x = 0 to x = 0.19
W = ½kx² from x = 0 to x = 0.19
W = ½k (0.19²-0²)
5 = ½k(0.0361-0)
5×2 = 0.0361k
Then, k = 10/0.0361
k = 277.008 N/m
The spring constant is 277.008N/m
Then, applying Hooke's law to find the applied force
F = kx
F = 277.008 × 0.19
F = 52.63 N
The applied force is 52.63N