Home / AP Physics 1: 2.3 Contact Forces- Exam Style questions with Answer- MCQ

AP Physics 1: 2.3 Contact Forces- Exam Style questions with Answer- MCQ

Question

Two identical blocks, block A and block B, are placed on different horizontal surfaces. Block A is initially at rest on a smooth surface, while block B is initially at rest on a rough surface. A constant horizontal force of magnitude F0 is exerted on each block. After the force has been applied for a time Δt, the speeds of blocks A and B are vA and vB, respectively. Which of the following claims indicates the correct relation between \(v_A\) and \(v_B\) and provides the best justification for the relation?

A \(v_A<v_B\). The forces between the atoms in a block and the atoms in a surface oppose the motion of the block and are greater, on average, for block A.

B \(v_A<v_B\). The forces between the atoms in a block and the atoms in a surface oppose the motion of the block and are greater, on average, for block B.

C \(v_A>v_B\). The forces between the atoms in a block and the atoms in a surface oppose the motion of the block and are greater, on average, for block A.

D \(v_A>v_B\). The forces between the atoms in a block and the atoms in a surface oppose the motion of the block and are greater, on average, for block B.

▶️Answer/Explanation

Ans:D

 The interatomic forces between the atoms in a block and the atoms in a surface oppose the motion of the block. If the horizontal force is exerted on each block, more force is required to move a block over a rough surface due to the microscopic ridges that impede the average movement of the block atoms. Accordingly, the speed of the block atoms moving over the smooth surface will be faster than the speed of the block atoms moving over the rough surface.

Question

An object is at rest on the ground. The object experiences a downward gravitational force from Earth. Which of the following predictions is correct about why the object does not accelerate downward? Select two answers. Justify your selections.

A The bonded molecules of the object are repelled upward by the bonded molecules of the ground with the same magnitude as the gravitational force downward on the object.

B The normal force is exerted upward on the object from the ground with the same magnitude as the gravitational force downward on the object.

C The bonded molecules of the object are attracted downward by the bonded molecules of the ground with the same magnitude as the gravitational force downward on the object.

D The force of friction is exerted upward on the object from the ground with the same magnitude as the gravitational force downward on the object.

▶️Answer/Explanation

Ans:A , B

The object and the ground are considered to be in direct contact. The object and the ground are composed of bonded particles that exist on the atomic level. The interaction of these particles results in an electric force that is exerted on the ground from the object and an electric force that is exerted on the object from the ground. Both electric forces are repulsive, are exerted on different objects, and are opposite in directions. For the object on the ground, an upward electric force is in the opposite direction as the gravitational force and has the same magnitude as the gravitational force on the object. This causes the object to be in static equilibrium.
A normal force can be explained as the electrostatic repulsion force on one object’s bonded molecules from another object’s bonded molecules as a result of the two objects essentially being in contact with each other. This upward electrostatic repulsion is in the opposite direction as the gravitational force and has the same magnitude as the gravitational force on the object. This causes the object to be in static equilibrium.

Question

A student uses an electronic force sensor to study how much force the student’s finger can apply to a specific location. The student uses one finger to apply a force on the sensor, and data collected from two trials are shown in the table. During which trial, if any, does the student’s finger experience the greatest electromagnetic force?

A Trial 1, because the student’s finger did not apply a force to the sensor.

B Trial 2, because the student’s finger applied the largest force to the sensor.

C An electromagnetic force is not exerted on the student’s finger for any trial because the sensor does not apply any force to the student’s finger.

D An electromagnetic force is not exerted on the student’s finger for any trial because the only force that the sensor applies to the student’s finger is a normal force.

▶️Answer/Explanation

Ans:B

Trial 2 is the trial in which the student exerted the greatest force on the sensor. Based on Newton’s third law of motion, the force that the student applies to the sensor will result in the sensor exerting a force of the same magnitude back onto the student’s finger. While the force that the sensor applies to the student’s finger is typically classified as a normal force, the normal force arises from the electromagnetic interactions of atoms and molecules at the atomic level that typically experience a repulsive force.

Question

A student pulls a wooden box along a rough horizontal floor at constant speed by means of a force P as shown to the right. Which of the following must be true?
(A) P > f and N < W.
(B) P > f and N = W.
(C) P = f and N > W.
(D) P = f and N = W.

▶️Answer/Explanation

Ans:A

Solution: Since P is at an upward angle, the normal force is decreased as P supports some of the weight. Since a component of P balances the frictional force, P itself must be larger than f.

Question

 A woman standing on a scale in an elevator notices that the scale reads her true weight. From this, she may conclude that
(A) the elevator must be at rest
(B) the elevator must be accelerating precisely at 9.8 \(m/s^{2}\)
(C) the elevator must be moving upward
(D) the elevator must not be accelerating

▶️Answer/Explanation

Ans:(D)

 The elevator must be traveling at constant velocity to insure the normal force (reading on the scale) is the same as her true weight. However, this velocity value can be any number: positive, zero, or negative.

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