An electron is located on the x axis at x0 = -9.31×10-6 m. Find the magnitude and direction of the electric field at x = 9.39 ×1
1 answer:
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Answer:
The distance is 11 m.
Explanation:
Given that,
Friction coefficient = 0.24
Time = 3.0 s
Initial velocity = 0
We need to calculate the acceleration
Using newton's second law
...(I)
Using formula of friction force
....(II)
Put the value of F in the equation (II) from equation (I)
....(III)
Put the value in the equation (III)
We need to calculate the distance,
Using equation of motion
Hence, The distance is 11 m.
Answer:
(a) W = 650J
(b) Wf = 529.2J
(c) W = 0J
(d) W = 0J
(e) ΔK = 120.8J
(f) v2 = 2.58 m/s
Explanation:
(a) In order to find the work done by the applied force you use the following formula:
(1)
F: applied force = 130N
d: distance = 5.0m
The work done by the applied force is 650J
(b) The increase in the internal energy of the box-floor system is given by the work done of the friction force, which is calculated as follow:
(2)
μ: coefficient of friction = 0.300
M: mass of the box = 36.0kg
g: gravitational constant = 9.8 m/s^2
The increase in the internal energy is 529.2J
(c) The normal force does not make work on the box because the normal force is perpendicular to the motion of the box.
(d) The same for the work done by the normal force. The work done by the gravitational force is zero because the motion of the box is perpendicular o the direction of the gravitational force.
(e) The change in the kinetic energy is given by the net work on the box. You use the following formula:
(3)
You calculate the total work:
(4)
F: applied force = 130N
Ff: friction force
d: distance = 5.00m
The friction force is:
Next, you replace the values of all parameters in the equation (4):
The change in the kinetic energy of the box is 120.8J
(e) The final speed of the box is calculated by using the equation (3):
(5)
v2: final speed of the box
v1: initial speed of the box = 0 m/s
You solve the equation (5) for v2:
The final speed of the box is 2.58m/s