Forces - MCAT Chemical and Physical Foundations of Biological Systems

Conservative forces are forces that do not lose energy to heat, sound, or light. Of these answers, energy is completely conserved and transferred from kinetic energy to potential energy, or vice versa. Gravitational forces, electrostatic forces, and elastic forces all work by providing a potential that will work in the same direction as the motion of an object or particle, allowing kinetic and potential energy to interconvert. Frictional forces lose energy as heat when sliding across a surface, and the more force (the more rough the surface), the more energy that is lost.

Friction is a non-conservative force, meaning that the work it does depends on the path taken by the object. For example, moving a brick in a long zig-zag across the table will generate more heat from friction than moving it in a straight line across the table.

Electric and gravitational forces are conservative. This can be tested by knowing a constant equation to calculate the energy associated with these forces; such equations are applicable regardless of path. No such equation exists for frictional energy.

The definition of mechanical advantage is:

Rearranging this equation and solving for weight gives:

Mechanical advantage is a unitless quantity. Remember the difference between weight and mass. Weight is a measure of force and has units of Newtons, whereas mass has units of kilograms.

Conservation of energy is the key here. Initial energy is all gravitational potential energy:

Note that the final height change is equal to the height above the spring added to the displacement of the spring.

This is equal to the final energy, which is all spring potential energy:

Set these equations equal and solve for the spring constant.

Choice 2 is correct. The formula for gravitational force, G, is , where k is a constant. The effect of doubling one mass and trebling the other is multiplicative, , so the answer is six-fold. The question attempts to confuse the respondent by forcing them to recall that the element of the equation which is squared is distance between objects.

You should be familiar with the following equation for the force of gravity.

To solve this problem, recognize that weight (Fg) is proportional to the inverse square of the radius. When the object was a distance of r (Earth’s radius) it had a weight of 500N. Now, the object is at a distance of 4r (radius of Earth plus the 3r distance that the object is moved to). With the proportion described above, we can see that the force is decreased by a factor of (4)2.

The law of gravitation is written as , with G being equal to .

Since the radius of the two masses acting on each other is squared, and is found in the denominator, a decrease in the radius by a multiple of three will cause a nine-fold increase in the gravitational force.

On earth, .

The law of universal gravitation is equal to .

We can set these equations equal to one another and isolate by dividing both sides by , the mass of an object on Earth.

Using the mass of the Earth, the radius of the Earth, and the gravitational constant, , we get a value of approximately if we solve for .

Gravity is essentially a property of mass, and the force of gravitational attraction between two bodies is given by the formula:

In our scenario, the masses remain the same, and of course is a constant, so the only thing that changes is the denominator.

We know , and the force of gravity is:

Also, the equation for uniform circular motion, such as a satellite in orbit is:

Set and substitute the acceleration due to circular motion into the equation. Solve for velocity.

This indicates that the only variables that affect the orbital speed are orbital radius and the mass of the Earth.

One way an object can be considered "weightless," is if it is accelerating downward at the same rate that it would be if it were free falling. This is why objects in an elevator whose cable had been cut would be "weightless." Therefore the answer is just for the block to be accelerating in the same magnitude and direction as the acceleration due to gravity, which is

In order for the potted plant to stay still, the net force on the plant must be 0N. The tension of the rope is the force that allows an object to hang without falling. In order to keep the plant stationary, the tension must be equal to 25N in the upward direction.

It is important to understand the difference between static and kinetic friction. When an object is at rest, it takes more force to get it to start moving than to keep it moving. If you match the amount of static friction that can be generated when the object is at rest, it will not move because there is zero net force; the force applied must be greater than the static friction in order to initiate motion. Once the object begins moving, the force required to keep it moving decreases. If you match the force of kinetic friction, the object moves at a constant velocity because there is again no net force. Any more force will cause acceleration, while any less will cause deceleration.

The skydiver’s velocity (along the y-axis) will be zero when he initially jumps out of the plane, however, his net force will be greatest at that point.

As the skydiver first jumps, there is only the force of gravity acting on him. As he approaches terminal velocity, however, the force of wind resistance becomes equal and opposite to the force of gravity, and he begins to move at constant velocity with a net force of zero.

Consider that tension is a contact force that acts over a distance. Assuming the rope does not stretch and has no mass (as we should for the MCAT), we can think of tension as a force that acts directly on the mass. We can draw a force diagram below.

If the system is stationary (at equilibrium), we can see that tension is equal to the weight, T = mg.

T = (2 kg)(9.8 m/s2) = 19.6N

Given that the system is motionless and that the tensional force is constant throughout the rope, we know that the weights on the platform must balance the weight of the mass. Given that the mass weights 2kg and each weight is 0.5kg, we can determine the number of weights required by dividing 2kg/0.5kg = 4 weights. Note that we can neglect the weight of the pulley in this problem; while this is usually the case on the MCAT as well, be sure to double check.

Use Newton's second law, , to relate the forces and accelerations of the particles. Writing this out for the two charges gives the following equations.

The two charges exert equal-magnitude and opposite-direction forces on one another according to Newton's third law, so , or using the right side of the second law equations, .

Plugging in our given values allows us to solve for .

The velocity of the mass will be the greatest at the equilibrium position. Although the restoring force on the mass due to the spring will be the greatest at maximum displacement from the equilibrium position, this force is being applied to the mass as it moves in the opposite direction. As the mass moves towards the equilibrium position from maximum displacement, the restoring force due to the spring is applied in the same direction as the velocity of the mass. Because of this, the object is accelerating. After the mass passes through the equilibrium position, the spring begins to apply a force in the direction opposite to the motion of the mass, resulting in the deceleration of the mass. The mass experiences its maximum velocity at the equilibrium position.

When an object is at terminal velocity, the upward force on the object from air resistance is equal to the force of gravity. These equal, but opposing, forces cancel each other out and the net force is zero. This results in a zero acceleration, allowing the object to fall at a constant (terminal) velocity.