Answer:
The work is -67.76 J
Explanation:
The law of conservation of energy is considered one of one of the fundamental laws of physics and states that the total energy of an isolated system remains constant. except when it is transformed into other types of energy.
This is summed up in the principle that energy can neither be created nor destroyed in the universe, only transformed into other forms of energy.
In this case you must calculate the loss of kinetic energy. This loss is actually the work done against the resistive force in the air. Friction is the only force other than gravity that acts on the ball.
So, the loss of kinetic energy is 
You know:
- mass=m=0.22 kg
- Initial velocity of the ball:

Final velocity of the ball: 
Replacing:
= -67.76 J
Friction work is always negative because friction is always against displacement.
<u><em>The work is -67.76 J</em></u>
Answer:
-20.158ft-lb
Explanation:
Check the attached files for the explanation.
Answer:
50 watts
Explanation:
Applying,
Power (P) = Workdone (W)/Time(t)
But,
Work done (W) = Force (F)×distance(d)
Therefore,
P = Fd/t..................... Equation 1
Where P = power of the weightlifter, F = Force applied, d = distance, t = time.
From the question,
Given: F = 200 N, d = 0.5 m, t = 2 s
Substitute these values into equation 1
P = (200×0.5)/2
P = 100/2
P = 50 watts
Answer:
a) 1504.8 J
b) 991.76 J
c) 0J
d) 0J
Explanation:
(a) The work done by the force P on the box is given by the following formula:

P: applied force = 171N
x: distance in which the for P is applied = 8.80m
you replace the values of P and x and obtain:

(b) The work don by the friction force is:

μ = coefficient of kinetic friction = 0.250
M: mass of the box = 46.0kg
g: gravitational constant = 9.8 m/s^2

(c) The Normal force is

but this force does not do work on the box because the direction is perpendicular to the direction of the force P.

(d) the same as before:

Answer:
Magnitude of the force on proton = F = 1.1085 × 10^-15 N
Explanation:
Charge on proton = q = 1.60 × 10^-19 C
Velocity of proton = V = 4.0 × 10^4 m/s
Magnetic field = B = 0.20 T
Angle between V and B = θ = 60
We know that,
F = qVBsin θ = (1.60 × 10^-19)( 4.0 × 10^4)( 0.20)sin(60)
F = 1.1085 × 10^-15 N