1. A 2.5 kg mass at rest on a horizontal surface is acted upon by an applied horizontal force of 150 N. A frictional force of 100 N acts on the mass. How much work is done on the mass?
my answer: would the work be 0?
2. 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: b
3.4. 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: b
A 2.5 kg mass at rest on a horizontal surface is acted upon by an applied horizontal force of 150 N. A frictional force of 100 N acts on the mass. How much work is done on the mass after 5 m?
First, we need to find the net force acting on the object:
Net force = Applied force - Frictional force
Net force = 150 N - 100 N
Net force = 50 N
Next, we can use the work-energy principle to find the work done on the mass:
Work = Change in kinetic energy
Since the mass is initially at rest, its initial kinetic energy is zero. We can find its final kinetic energy using the work done by the net force over the distance of 5 m:
Work = Force x Distance x cos(theta)
Work = 50 N x 5 m x cos(0)
Work = 250 J
Final kinetic energy = (1/2) x mass x velocity^2
Since the mass is brought to a stop, its final velocity is zero. Therefore, its final kinetic energy is also zero.
Change in kinetic energy = Final kinetic energy - Initial kinetic energy
Change in kinetic energy = 0 - 0
Change in kinetic energy = 0
So, the work done on the mass is 250 J.
A 2.5 kg mass at rest on a horizontal surface is acted upon by an applied horizontal force of 150 N. A frictional force of 100 N acts on the mass. How much work is done on the mass after 5 m
The net force acting on the mass is:
Net force = Applied force - Frictional force
Net force = 150 N - 100 N
Net force = 50 N
The work done by this net force over a distance of 5 m is:
Work = Net force x Distance x cos(theta)
Work = 50 N x 5 m x cos(0)
Work = 250 J
Therefore, the work done on the mass after 5 m is 250 J.
A mass of 10 kg, initially at rest on a horizontal frictionless surface, is acted upon by a horizontal force of 25 N. The speed of the mass after it has moved 8.0 m is
We can begin by using Newton's second law and the work-energy principle to solve this problem.
Newton's second law tells us that the net force on an object is equal to its mass times its acceleration:
F_net = m*a
In this case, the net force on the mass is the applied force of 25 N, since there is no frictional force:
F_net = 25 N
So, we can find the acceleration of the mass using:
a = F_net/m
a = 25 N / 10 kg
a = 2.5 m/s^2
Next, we can use the work-energy principle to find the final velocity of the mass after it has moved 8.0 m:
Work = Change in kinetic energy
The work done on the mass is equal to the force applied times the distance it moves:
Work = Force x Distance x cos(theta)
Work = 25 N x 8.0 m x cos(0)
Work = 200 J
The initial kinetic energy of the mass is zero since it is at rest. The final kinetic energy can be found using:
Kinetic energy = (1/2)*m*v^2
where v is the final velocity of the mass. We can set the work done equal to the change in kinetic energy:
Work = Final kinetic energy - Initial kinetic energy
200 J = (1/2)*10 kg*v^2 - 0
Solving for v, we get:
v = sqrt(200 J / (1/2*10 kg))
v = sqrt(400)
v = 20 m/s
Therefore, the speed of the mass after it has moved 8.0 m is 20 m/s.
A 100 kg mass is dropped from a height of 3.0 m. The kinetic energy before striking the ground is:
We can use the work-energy principle to solve this problem. The potential energy of the mass at the initial height is converted entirely into kinetic energy as it falls, ignoring air resistance and any other forms of energy loss.
The potential energy of the mass at a height of 3.0 m can be found using:
Potential energy = m*g*h
where m is the mass of the object, g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the object.
Potential energy = 100 kg * 9.8 m/s^2 * 3.0 m
Potential energy = 2940 J
According to the work-energy principle, the potential energy is converted into kinetic energy as the object falls. Therefore, the kinetic energy just before striking the ground is equal to the potential energy at the initial height:
Kinetic energy = 2940 J
So, the kinetic energy just before striking the ground is 2940 J.
A 300 N force is applied horizontally to a 50 kg crate pushing it 2 m. The force of friction between the crate and the floor is 200 N. The magnitude of the work done by the applied force is:
The net force acting on the crate is:
Net force = Applied force - Frictional force
Net force = 300 N - 200 N
Net force = 100 N
So, the net force acting on the crate is 100 N. We can use the work-energy principle to find the work done by this net force over a distance of 2 m:
Work = Change in kinetic energy
The initial kinetic energy of the crate is zero since it is at rest. The final kinetic energy can be found using:
Kinetic energy = (1/2)*m*v^2
where m is the mass of the crate and v is its final velocity. We can solve for v using the equations of motion:
v^2 = u^2 + 2*a*s
where u is the initial velocity (zero), a is the acceleration, and s is the distance traveled. Since the net force acting on the crate is constant, the acceleration is also constant:
a = Net force / mass
a = 100 N / 50 kg
a = 2 m/s^2
Using this acceleration, we can find the final velocity of the crate:
v^2 = 0 + 2*2 m/s^2 * 2 m
v^2 = 8 m^2/s^2
v = sqrt(8) m/s
Therefore, the final velocity of the crate is 2.83 m/s.
The change in kinetic energy is:
Change in kinetic energy = (1/2)*m*v^2 - 0
Change in kinetic energy = (1/2)*50 kg*(2.83 m/s)^2
Change in kinetic energy = 199.6 J
So, the magnitude of the work done by the applied force is 199.6 J.
A 300 N force is applied horizontally to a 50 kg crate pushing it 2 m. The force of friction between the crate and the floor is 200 N. The magnitude of the work done by the frictional force is
The frictional force acts opposite to the direction of motion of the crate, so it does negative work. The work done by the frictional force can be found using the same equation as for the work done by the net force:
Work = Change in kinetic energy
In this case, we can find the change in kinetic energy using the final velocity of the crate found in the previous question:
Change in kinetic energy = (1/2)*m*v^2 - 0
Change in kinetic energy = (1/2)*50 kg*(2.83 m/s)^2
Change in kinetic energy = 199.6 J
Since the net work done on the crate is equal to the sum of the work done by the applied force and the work done by the frictional force, we can rearrange the equation to find the work done by the frictional force:
Work done by frictional force = Net work - Work done by applied force
Work done by frictional force = Change in kinetic energy - Work done by applied force
Work done by frictional force = 199.6 J - 300 J
Work done by frictional force = -100.4 J
The negative sign indicates that the work done by the frictional force is in the opposite direction to the displacement of the crate. Therefore, the magnitude of the work done by the frictional force is 100.4 J.
A 1000 kg car whose speed is 80 km/h has a kinetic energy o
We can use the formula for kinetic energy to solve this problem:
Kinetic energy = (1/2) * mass * velocity^2
where mass is the mass of the car and velocity is its speed. We need to convert the speed from km/h to m/s, first:
80 km/h = 80,000 m/3600 s = 22.22 m/s
Now we can plug in the values to find the kinetic energy:
Kinetic energy = (1/2) * 1000 kg * (22.22 m/s)^2
Kinetic energy = 1/2 * 1000 * 493.8
Kinetic energy = 246900 J
Therefore, the kinetic energy of the car is 246900 J.
A girl wants to slide down a frictionless playground slide. If she starts at rest from 'X' and slides down to 'Y', which of the following slide shapes will give the girl the greatest possible speed upon reaching 'Y'? All the slides have the same vertical height and THEY ARE FRICTIONLESS.
The height of the slide is constant, so the final potential energy of the girl is the same for all slide shapes. However, the final kinetic energy of the girl will depend on her final speed, which is determined by the slide shape.
According to the law of conservation of energy, the total mechanical energy of the girl is conserved as she slides down the slide. Since the slide is frictionless, the total mechanical energy is equal to the initial potential energy of the girl:
Potential energy at X = Potential energy at Y
mgh = (1/2)mv^2
where m is the mass of the girl, g is the acceleration due to gravity, h is the height of the slide, and v is the final speed of the girl.
We can rearrange this equation to solve for v:
v = sqrt(2gh)
The speed of the girl at the bottom of the slide is proportional to the square root of the height of the slide. The higher the slide, the faster she will be going when she reaches the bottom.
Therefore, the slide shape that will give the girl the greatest possible speed upon reaching 'Y' is the slide with the steepest slope. A steeper slope will result in a greater change in height over a shorter distance, leading to a greater final speed.
A jet engine applies a force of 300 N horizontally to the right against a 50 kg rocket. The force of air friction 200 N. If the rocket is thrust horizontally 2.0 m, the kinetic energy gained by the rocket is
The net force acting on the rocket is:
Net force = Thrust force - Frictional force
Net force = 300 N - 200 N
Net force = 100 N
We can use the work-energy principle to find the kinetic energy gained by the rocket as it moves 2.0 m due to the net force acting on it:
Work done on the rocket = Change in kinetic energy
The work done on the rocket is equal to the force acting on it times the distance it moves:
Work done on the rocket = Net force x Distance x cos(theta)
Work done on the rocket = 100 N x 2.0 m x cos(0)
Work done on the rocket = 200 J
The initial kinetic energy of the rocket is zero since it is at rest. The final kinetic energy of the rocket after moving 2.0 m due to the net force is:
Final kinetic energy = (1/2)*mass*velocity^2
We can find the final velocity of the rocket using the equations of motion:
v^2 = u^2 + 2*a*s
where u is the initial velocity (zero), a is the acceleration of the rocket, and s is the distance it moves.
The acceleration of the rocket can be found using the net force and the mass of the rocket:
Acceleration = Net force / mass
Acceleration = 100 N / 50 kg
Acceleration = 2 m/s^2
Substituting the values into the equations of motion:
v^2 = 0 + 2 * 2 m/s^2 * 2.0 m
v^2 = 8 m^2/s^2
Therefore, the final velocity of the rocket is:
v = √(8 m^2/s^2)
v = 2.83 m/s
Now, we can find the final kinetic energy of the rocket:
Final kinetic energy = (1/2)*mass*velocity^2
Final kinetic energy = (1/2)*50 kg*(2.83 m/s)^2
Final kinetic energy = 199.6 J
Therefore, the kinetic energy gained by the rocket is 199.6 J.
A jet engine applies a force of 300 N horizontally to the right against a 50 kg rocket. The force of air friction 200 N. If the rocket is thrust horizontally 2.0 m, the velocity of the rocket is
We can use the work-energy principle to relate the work done by the net force to the change in kinetic energy of the rocket:
Work done by net force = Change in kinetic energy
The net force on the rocket is the difference between the thrust force and the frictional force:
Net force = Thrust force - Frictional force
Net force = 300 N - 200 N
Net force = 100 N
The work done by the net force over a displacement of 2.0 m is:
Work done by net force = Net force x Distance x cos(theta)
Work done by net force = 100 N x 2.0 m x cos(0)
Work done by net force = 200 J
The initial kinetic energy of the rocket is zero since it is at rest. Therefore, the final kinetic energy is equal to the work done by the net force:
Final kinetic energy = Work done by net force
Final kinetic energy = 200 J
We can use the equation for kinetic energy to find the final velocity of the rocket:
Final kinetic energy = (1/2) * mass * velocity^2
Substituting the values:
200 J = (1/2) * 50 kg * velocity^2
Solving for velocity:
velocity^2 = (2 * 200 J) / 50 kg
velocity^2 = 8 m^2/s^2
Therefore, the velocity of the rocket is:
velocity = sqrt(8 m^2/s^2)
velocity = 2.83 m/s
A 100 kg mass is dropped from a height of 3.0 m. The kinetic energy before striking the ground is:
The potential energy of the mass at a height of 3.0 m can be found using:
Potential energy = m * g * h
where m is the mass of the object, g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the object.
Potential energy = 100 kg * 9.8 m/s^2 * 3.0 m
Potential energy = 2940 J
According to the work-energy principle, the potential energy is converted into kinetic energy as the object falls. Therefore, the kinetic energy just before striking the ground is also 2940 J.
So, the kinetic energy before striking the ground is 2940 J.
1. A 2.5 kg mass at rest on a horizontal surface is acted upon by an applied horizontal force of 150 N. A frictional force of 100 N acts on the mass. How much work is done on the mass?
my answer: would the work be 0?
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Yes, no motion, no work
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2. 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: b
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NO ! m g the same all the time
I say a) m g h + (1/2) m v^2 = constant
h goes up, v goes down
3.4. 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: b
NO !
(1/2) m v^2 and (1/2)m(2 v)^2
2^2 = 4
Ke of second four times the first
therefore 4 times as much work to stop
so (c)