A system of two objects has ΔKtot = 6 J and ΔUint = -5 J. Part A How much work is done by interaction forces? Express your answe
1 answer:
A) +5 J
B) +1 J
Explanation:
A)
The internal forces (interaction forces) acting on a system do not change the mechanical energy (sum of potential and kinetic energy) of the system.
However, these forces are responsible for converting the energy from one form into another; the work done by these forces is equal to the amount of energy converted from one form into the other.
In this problem, we have:
is the loss in potential energy of the system
is the gain in kinetic energy of the system
By looking at these numbers, this means that the internal forces have converted 5 J of energy from potential energy into kinetic energy (while the additional +1 J missing is due to external forces, as explained in part B).
Therefore, the work done by internal forces is
W = +5 J
B)
First of all, we calculate the change in mechanical energy of the system.
The mechanical energy of a system is the sum of its kinetic energy (K) and its potential energy (U):
So, the change in mechanical energy is equal to the sum of the changes of kinetic energy and the changes of potential energy:
In this problem:
So, the change in mechanical energy is:
According to the work-energy theorem, the work done by external forces on a system is equal to the change in mechanical energy of the system: therefore in this case, the work done by external forces is
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a) 2.64 s
We can solve this part of the problem by using the following SUVAT equation:
where
s is the displacement of the stone
u is the initial velocity
t is the time
a is the acceleration
We must be careful to the signs of s, u and a. Taking upward as positive direction, we have:
- s (displacement) negative, since it is downward: so s = -75.0 m
- u (initial velocity) positive, since it is upward: +15.5 m/s
- a (acceleration) negative, since it is downward: so a= g = -9.8 m/s^2 (acceleration of gravity)
Substituting into the equation,
Solving the equation, we have two solutions: t = -5.80 s and t = 2.84 s. Since the negative solution has no physical meaning, the stone reaches the bottom of the cliff 2.64 s later.
b) 10.4 m/s
The speed of the stone when it reaches the bottom of the cliff can be calculated by using the equation:
where again, we must be careful to the signs of the various quantities:
- u (initial velocity) positive, since it is upward: +15.5 m/s
- a (acceleration) negative, since it is downward: so a = g = -9.8 m/s^2
Substituting t = 2.64 s, we find the final velocity of the stone:
where the negative sign means that the velocity is downward: so the speed is 10.4 m/s.
c) 4.11 s
In this case, we can use again the equation:
where
s is the displacement of the package
u is the initial velocity
t is the time
a is the acceleration
We have:
s = -105 m (vertical displacement of the package, downward so negative)
u = +5.40 m/s (initial velocity of the package, which is the same as the helicopter, upward so positive)
a = g = -9.8 m/s^2
Substituting into the equation,
Which gives two solutions: t = -5.21 s and t = 4.11 s. Again, we discard the first solution since it is negative, so the package reaches the ground after
t = 4.11 seconds.
Answer:
Fp = 26.59[N]
Explanation:
This problem can be solved using the principle of work and energy conservation, i.e. the final kinetic energy of a body will be equal to the sum of the forces that do work on the body plus the initial kinetic energy.
We need to identify the initial data:
d = distance = 41.9[m]
Ff = friction force = 44.5 [N]
m = mass = 16.3 [kg]
v1 = 1.9 [m/s]
v2 = 12.6 [m/s]
The kinetic energy at the beginning can be calculated as follows:
And the final kinetic energy.
The work is performed by two forces, the friction force and the pushing force, it is important to clarify that these forces are opposite in direction.
The weight of the cart also performs a work in the direction of movement since the plane is tilted down, this component of the weight of the cart must be parallel to the surface of the inclined plane.