Question

a box weighing 29 n is pulled horizontally until it slides uniformly over a level floor. if the applied force is 6.8 n, what is the coefficient of friction between the box and the floor?
0.42
0.67
0.23
1.23

if a box has a mass of 6.0 kg, what will the normal force of the table on the box be?
67 n
59 n
82 n
9.8 n

Explanation:

First Question (Coefficient of Friction)

Step1: Recall the friction formula

When an object slides uniformly (constant velocity), the applied force \( F_{applied} \) equals the frictional force \( F_f \). The formula for frictional force is \( F_f=\mu N \), where \( \mu \) is the coefficient of friction and \( N \) is the normal force. On a level floor, the normal force \( N \) equals the weight of the object, so \( N = 29\space N \) (given weight of the box). The applied force \( F_{applied}=6.8\space N \), so \( F_f = 6.8\space N \).

Step2: Solve for \( \mu \)

From \( F_f=\mu N \), we can rearrange to solve for \( \mu \): \( \mu=\frac{F_f}{N} \). Substituting the values \( F_f = 6.8\space N \) and \( N = 29\space N \), we get \( \mu=\frac{6.8}{29}\approx0.23 \).

Second Question (Normal Force)

Step1: Recall the normal force on a level surface

On a level surface, the normal force \( N \) equals the weight of the object. The weight \( W \) is given by \( W = mg \), where \( m = 6.0\space kg \) and \( g = 9.8\space m/s^2 \) (acceleration due to gravity).

Step2: Calculate the weight (normal force)

Substitute \( m = 6.0\space kg \) and \( g = 9.8\space m/s^2 \) into \( W = mg \): \( W=(6.0)(9.8)=58.8\space N\approx59\space N \). So the normal force is approximately 59 N.

Answer:

0.23