F = M A
Force = (mass) x (acceleration)
= (1,650 kg) (4 m/s²) = 6,600 kg-m/s² = <em>6,600 Newtons</em>
Answer:
v = 0.41 m/s
Explanation:
- In this case, the change in the mechanical energy, is equal to the work done by the fricition force on the block.
- At any point, the total mechanical energy is the sum of the kinetic energy plus the elastic potential energy.
- So, we can write the following general equation, taking the initial and final values of the energies:
![\Delta K + \Delta U = W_{ffr} (1)](https://tex.z-dn.net/?f=%5CDelta%20K%20%2B%20%5CDelta%20U%20%3D%20W_%7Bffr%7D%20%20%281%29)
- Since the block and spring start at rest, the change in the kinetic energy is just the final kinetic energy value, Kf.
- ⇒ Kf = 1/2*m*vf² (2)
- The change in the potential energy, can be written as follows:
![\Delta U = U_{f} - U_{o} = \frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (3)](https://tex.z-dn.net/?f=%5CDelta%20U%20%3D%20U_%7Bf%7D%20%20-%20U_%7Bo%7D%20%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%2A%20k%20%2A%20%28x_%7Bf%7D%20%5E%7B2%7D%20-%20x_%7B0%7D%20%5E%7B2%7D%20%29%20%283%29)
where k = force constant = 815 N/m
xf = final displacement of the block = 0.01 m (taking as x=0 the position
for the spring at equilibrium)
x₀ = initial displacement of the block = 0.03 m
- Regarding the work done by the force of friction, it can be written as follows:
![W_{ffr} = - \mu_{k}* F_{n} * \Delta x (4)](https://tex.z-dn.net/?f=W_%7Bffr%7D%20%3D%20-%20%5Cmu_%7Bk%7D%2A%20F_%7Bn%7D%20%2A%20%5CDelta%20x%20%20%284%29)
where μk = coefficient of kinettic friction, Fn = normal force, and Δx =
horizontal displacement.
- Since the surface is horizontal, and no acceleration is present in the vertical direction, the normal force must be equal and opposite to the force due to gravity, Fg:
- Fn = Fg= m*g (5)
- Replacing (5) in (4), and (3) and (4) in (1), and rearranging, we get:
![\frac{1}{2} * m* v^{2} = W_{ffr} - \Delta U = W_{ffr} - (U_{f} -U_{o}) (6)](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20%2A%20m%2A%20v%5E%7B2%7D%20%3D%20W_%7Bffr%7D%20-%20%5CDelta%20U%20%3D%20W_%7Bffr%7D%20-%20%28U_%7Bf%7D%20-U_%7Bo%7D%29%20%20%286%29)
![\frac{1}{2} * m* v^{2} = (- \mu_{k}* m*g* \Delta x) -\frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (7)](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20%2A%20m%2A%20v%5E%7B2%7D%20%3D%20%28-%20%5Cmu_%7Bk%7D%2A%20m%2Ag%2A%20%5CDelta%20x%29%20%20-%5Cfrac%7B1%7D%7B2%7D%20%2A%20k%20%2A%20%28x_%7Bf%7D%20%5E%7B2%7D%20-%20x_%7B0%7D%20%5E%7B2%7D%20%29%20%287%29)
- Replacing by the values of m, k, g, xf and x₀, in (7) and solving for v, we finally get:
![\frac{1}{2} * 2.00 kg* v^{2} = (-0.4*2.00 kg*9.8m/s2*0.02m) +( (\frac{1}{2} *815 N/m)* (0.03m)^{2} - (0.01m)^{2}) = -0.1568 J + 0.326 J (8)](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20%2A%202.00%20kg%2A%20v%5E%7B2%7D%20%20%3D%20%28-0.4%2A2.00%20kg%2A9.8m%2Fs2%2A0.02m%29%20%2B%28%20%28%5Cfrac%7B1%7D%7B2%7D%20%2A815%20N%2Fm%29%2A%20%280.03m%29%5E%7B2%7D%20-%20%280.01m%29%5E%7B2%7D%29%20%3D%20-0.1568%20J%20%2B%200.326%20J%20%288%29)
I believe that the statement above is TRUE. I can say that it's true because based on the definition of what reciprocal liking is, this is the probability of having to like someone who also likes them in return. Therefore, if they like each other, this means that there is a low risk of rejection in the relationship.
<span>The bright, visible surface of the Sun is called corona. The outermost layer of the Sun's atmosphere is called chromosphere.</span>