Question

Figure 6-20 shows an initially stationary block of mass m on a floor. A force of magnitude 0.500mg is then applied at upward angle ? = 20°. What is the magnitude of the acceleration of the block across the floor if (a)µs = 0.620 and µk = 0.540 and (b)µs = 0.420 and µk = 0.330?

Answer 1

Concepts and reason

The concepts used to solve the problem are Newton’s second law of motion, kinetic friction, and static friction.

Initially, use Newton’s second law, force of static friction, and horizontal component of force to find the acceleration.

Finally, use Newton’s second law, force of kinetic friction, and coefficient of kinetic friction to find the acceleration once the block starts to move.

Fundamentals

Newton’s second law states that “For a body, the rate at which its momentum changes is proportional to the net force acting on the body”.

The expression for Newton’s second law is as follows:

Here, the net force acting on a body is , the mass of the body is , and the acceleration produced on it is .

The expression for the force of static friction is given below:

Here, the force of static friction is , the coefficient of static friction , and the normal force is .

The expression for the force of kinetic friction is as follows:

Here, the force of kinetic friction is , and the coefficient of kinetic friction is .

(a)

The free body diagram is represented below:

Here, angle of force with the horizontal is , and the acceleration due to gravity is .

The force is resolved into in the horizontal direction and in the vertical direction. The force of gravity acts downward and the normal force acts opposite to the force of gravity.

The expression for Newton’s second law is as follows:

From the free body diagram, the net force acting along the vertical direction is as follows:

Rearrange the above equation to obtain the normal force.

Substitute for and for .

The force of static friction acting on the cube to get sliding is given below:

Substitute for and for .

The forces acting along the horizontal direction are , and the force of static friction acts opposite to the direction of motion.

Thus, the net force acting on the cube along the horizontal direction is as follows:

Substitute for and for .

Here, . Hence, the block remains stationary. The acceleration of the block is

(b)

The expression for Newton’s second law is as follows:

From the free body diagram, the net force acting along the vertical direction is as follows:

Rearrange the above equation to obtain the normal force.

Substitute for and for .

The force of static friction acting on the cube to get sliding is given below:

Substitute for and for .

The force acting along the horizontal direction is , and the force of static friction acts opposite to the direction of motion.

Substitute for and for .

Here, . Hence, the block has acceleration.

The force of kinetic friction acting on the cube while moving is as follows:

Substitute for .

From Newton’s second law, the expression for net force is as follows:

Here, the net force is .

The force acting on the block once it starts to move is , and the force of kinetic friction acts opposite to each other.

Thus, the net force acting on the block once it starts to move is given below:

Substitute for and for .

Rearrange the above equation to obtain acceleration.

Substitute for , for , for , and for .

Ans: Part a

The magnitude of acceleration of the block across the floor is .

Part b

The magnitude of acceleration of block across the floor is .