Chapter 6, Problem 4Q

Can the normal force on an object ever do work? Explain.

To determine

Whether the normal force on an object ever do work.

Answer to Problem 4Q

Solution:The normal force cannot do work on an object.

Explanation of Solution

Introduction: Work done $(W)$ by a net force $\left({\overrightarrow{F}}_{\text{net}}\right)$ is defined as:

  $W={\overrightarrow{F}}_{\text{net}}𑨀\Delta \overrightarrow{x}=\left|{\overrightarrow{F}}_{\text{net}}\right|.\left|\Delta \overrightarrow{x}\right|.\cos (\theta )$  (1)

Where:

• $\left|{\overrightarrow{F}}_{\text{net}}\right|$ isthe magnitude of the net force ${\overrightarrow{F}}_{\text{net}}$ .
• $\left|\Delta \overrightarrow{x}\right|$ is the magnitude of displacement $\Delta \overrightarrow{x}$ of the object.
• $\theta$ is the angle between net force ${\overrightarrow{\text{F}}}_{\text{net}}$ and displacement $\Delta \overrightarrow{x}$ .

Suppose that anobject on a surface tableis pushed to the right with a force $\left(\text{<mover accent="true"><mo>F</mo><mo>→</mo></mover>}\right)$ . Now, the work done $\left(W_{\text{N}}\right)$ by the normal force $\left(\text{<mover accent="true"><mo>N</mo><mo>→</mo></mover>}\right)$

according to equation (1) is:

  $W_N=\left|\overrightarrow{N}\right|\left|\Delta \overrightarrow{x}\right|.\cos (\theta )$  (2)

But the normal force is perpendicular to displacement $\left(\Delta \overrightarrow{x}\right)$ , so:

  $\theta =90°$  (3)

Substitute (3) into equation (2):

  $W_N=0$  (4)

Even if the object moves in the same direction as the normal force, as soon as the object leaves the contact from the surface, normal force becomes zero because normal force is a contact force.

Conclusion:

According to equation (4), the normal force can’t do work on an object.