1. In moving an object of mass 10 kg through a distance of 8.0 m, 40 J of work is done. The average force exerted is?
my answer: 320N
2. What is the kinetic energy of a 20 kg object moving at a speed of 10 m/s?
my answer: 1000J
3. the gravitational potential energy of a bird of mass 2.0 kg on a tree branch of height 15 m is most nearly
my answer: 300J
4. Two objects have the same mass. One is travelling twice as fast as the other. The work that must be done to stop the faster object compared to the work required to stop the slower object is:
a)two times greater
b)the same
c)four times greater
d)half as great
e)one quarter as great
my answer: a
5. When a rock is thrown straight up in the air, after it leaves the hand, the rock begins to slow down. This occurs because:
a) the rock is gaining potential energy as it rises and thus it must lose kinetic energy
b)the force of gravity acting on the rock increases as the rock rises
c)the forces acting on the rock are balanced
d)the potential energy of the rock decreases as the rock rises
my answer: d
I believe that for 1 the answer is 5N. Work = Force * Distance and since we have two of those variables we can figure out the other one. We can substitute 40 J for W and 8 m for D.
40J = F * 8m
Then we isolate F so we can solve for it.
40J/8m = F
5N = F
In moving an object of mass 10 kg through a distance of 8.0 m, 40 J of work is done. The average force exerted is
5N.
What is the kinetic energy of a 20 kg object moving at a speed of 10 m/s?
The kinetic energy (KE) of an object with mass (m) moving at a velocity (v) is given by the equation:
KE = 1/2 * m * v^2
Plugging in the given values, we get:
KE = 1/2 * 20 kg * (10 m/s)^2
KE = 1000 J
Therefore, the kinetic energy of the 20 kg object moving at 10 m/s is 1000 J.
The gravitational potential energy of a bird of mass 2.0 kg on a tree branch of height 15 m is closest to
In moving an object of mass 10 kg through a distance of 8.0 m, 40 J of work is done. If the kinetic energy gained by the mass is 34 J what is the WORK DONE by the NON CONSERVATIVE force?
We can solve for the work done by the non-conservative force using the work-energy principle, which states that the net work done on an object is equal to its change in kinetic energy. Thus, we have:
Net work = Change in kinetic energy
Total work - Non-conservative work = Kinetic energy gained
40 J - Non-conservative work = 34 J
Non-conservative work = 6 J
Therefore, the work done by the non-conservative force is 6 J.
In moving an object of mass 10 kg through a distance of 8.0 m, 40 J of work is done. If the kinetic energy gained by the mass is 34 J what is the EFFICIENCY?
The efficiency (η) is defined as the ratio of the output energy to the input energy, expressed as a percentage. In this case, the output energy is the kinetic energy gained by the mass (34 J), and the input energy is the work done on the mass (40 J). Thus:
η = (output energy / input energy) x 100%
η = (34 J / 40 J) x 100%
η = 0.85 x 100%
η = 85%
Therefore, the efficiency is 85%.