Test: Force and Newton`s laws of Motion - MCAT MCQ

The First Law states that an object at rest stays at rest and an object in motion (at constant velocity) stays in motion unless an outside (unbalanced) force acts on it.

So an object with zero velocity and at a constant velocity are considered similar states, not acceleration.

Mass is a measure of inertia, which is the ability to resist motion. An object with a large mass would be difficult to stop or get going, not speed.

The tendency to continue in a straight line is an example of inertia.

Zero acceleration means that there is no net force acting on the object according to Newton’s second law.

Zero net force does not mean there are no forces acting on the object. If there is a force, there must be an equal and opposite force to balance it.

Zero acceleration could mean two possible scenarios: zero velocity or constant velocity, and, consequently, motionless or in motion.

What must be true is that the object has no unbalanced forces acting on it.

Uniform circular motion is characterized by motion in a circular path at a constant speed due to a centripetal force towards an axis of rotation.

The object does not move with constant velocity since velocity constantly changes in direction but remains the same in magnitude.

The two vectors, velocity and acceleration, are not tangential but perpendicular to one another.

Objects in uniform circular motion are subject to a center-seeking, centripetal, force.

If the acceleration vector went to zero or there were no centripetal force, the object would not deviate in direction, but travel in straight line at a constant velocity.

• Acceleration is directly proportional to force and inversely proportional to mass: F = ma.
• If mass is doubled, acceleration would be halved: F = 2m(½a).
• If force is reduced by four, acceleration would be reduced by four: ¼F = m(¼a).
• In total, the acceleration decreases by a factor of 8: ¼F = 2m(⅛a) = ¼(ma).

• An unbalanced force would cause an acceleration according to Newton’s second law. Change in acceleration can be due to either change in magnitude or direction.
• The object would not be stationary or traveling at a constant velocity.
• Motion in a straight line eliminates the possibility of change in direction, but there may be change in magnitude or not.
• When an object changes direction and travels with the same speed it exhibits an acceleration, in which there must be an unbalanced force.

• An action-reaction pair must consist of forces that are of the same type (field or contact) and that act on different objects.
• Weight is a field force, and normal force is a contact force, so they are not of the same type and act on the same object, i.e. book.
• At terminal velocity, the force of gravity equals the force of air resistance, but they are not the same type and act on the same object, i.e. falling object.
• Gravity along the plane of the incline and the static frictional force are equal, but they are not the same type and act on the same object, i.e. block.
• The only action-reaction pair is the force of the horse on the cart and force of the cart on the horse.

• Whether the convertible exerts a smaller force because of its smaller mass cannot be determined since it has a greater change in velocity and therefore, acceleration to offset its smaller mass.
• Whether the convertible exerts a larger force because of its greater velocity cannot be determined since the truck has a larger mass despite its smaller velocity. It would take more work to stop the convertible due its higher velocity.
• It is the energy of the collision that gets dissipated in the form of heat, sound, and deformation of the two vehicles, not the force.
• They exert the same force on each other is correct. Due to Newton’s Third Law, if the truck collides with force F, then the SUV collides with force -F, equal in magnitude opposite in direction.
• Additionally, their momentums are equal, so that their impulses are equal. Therefore, the time of the collision is the same for both, so the average force of impact is the same.

The formula for center of mass is reminiscent of the equation for conservation of momentum for totally inelastic collisions, except that x replaces v.

Find the center of mass for his first position:

Find the center of mass for his second position. We assume the boat has shifted a distance x’ in the direction opposite the motion of the scientist. If we are wrong then we will get a negative value for x’.
 

The boat starts in equilibrium, and will remain in equilibrium throughout the walk, so the center of mass remains stationary. Therefore we can set the expressions equal to one another to solve for x’:

The correct answer is that the center of mass shifts by L/4 in the opposite direction of the scientist’s motion.

• If the force acts downwards to the left or upwards to the right of the fulcrum, then there is a counterclockwise (CCW) torque. If the force acts downwards to the right or upwards to the left of the fulcrum, then there is a clockwise (CW) torque.
• Forces will act from the center of mass if possible, and the fulcrums are the top right edge of Block 2 and the left edge of the desk.
• If Block 2 were shifted to the left, the center of mass moves to the left of the fulcrum, so the weight creates a CCW torque, not CW. The weight of Block 1 acts to the right of the same fulcrum downwards creating a CW torque, not CCW.
• If Block 1 were shifted to the right, as the center of mass moves to the right of the fulcrum, it will create a CW torque that topples off Block 1. However, the small normal force from Block 2 has been decreasing as Block 1 moved to the right.
• The weight of Block 1 acts to the left of the fulcrum downwards, and the weight will produce a CCW torque, not CW.
• There is a normal force from Block 2 on Block 1 that acts to the left of the fulcrum upwards, producing a CW torque.
• Here is a diagram indicating the forces:
  

• All forces need to be resolved along the plane of the ladder such that it becomes the x-axis and the y-axis perpendicular to that, like on an inclined plane.
• Since the fulcrum is placed at the bottom of the ladder and F_NG acts at the fulcrum, it is not considered to produce a torque.
• F_NW points directly outward from the wall to the left. F_NW cosθ is along the plane of the ladder, so F_NW sin θ is the force that would create a positive or counterclockwise torque on the ladder.
• F_FW points from the contact point of the ladder on the wall upwards along the wall. F_NW sin θ is the force component that is along the plane of the ladder pointing into the wall. F_FW cos θ would be the force that creates a counterclockwise torque.
• W points downward from the center of the ladder. Wcosθ is perpendicular to the plane of the ladder and creates a clockwise torque on the ladder.
• Here is a diagram illustrating all the forces creating the torque: