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.