Practice #1

A n applied force of 50 N is used to accelerate an object to the right across a frictio-nal surface. The object encounters 10 N of friction. Use the diagram to determine the normal force, the net force, the mass, and the acceleration of the object. (Neg-lect air resistance.)

N ote: To simplify calculations, an approximated value of g is often used - 10 m/s/s. Answers obtained using this approximation are shown in parenthesis.

F_norm = 80 N; m = 8.16 kg; F_net = 40 N, right; a = 4.9 m/s/s, right

( F_norm = 80 N; m = 8 kg; F_net = 40 N, right; a = 5 m/s/s, right )

Since there is no vertical acceleration, normal force = gravity force.

The mass can be found using the equation Fgrav = m * g.

The Fnet is the vector sum of all the forces: 80 N, up plus 80 N, down equals 0 N. And 50 N, right plus 10 N, left = 40 N, right. Finally, a = Fnet / m = (40 N) / (8.16 kg) = 4.9 m/s/s.

Practice #2

An applied force of 20 N is used to accelerate an object to the right across a frictional surface. The object encounters 10 N of friction. Use the diagram to determine the normal force, the net force, the coefficient of friction (μ) between the object and the surface, the mass, and the acceleration of the object. ( Neglect air resistance .)

Note: To simplify calculations, an approximated value of g is often used - 10 m/s/s. Answers obtained using this approximation are shown in parenthesis.

 

F_norm = 100 N; m = 10.2 kg; F_net = 10 N, right; "mu" = 0.1; a =0.980 m/s/s, right

( F_norm = 100 N; m = 10 kg; F_net = 10 N, right; "mu" = 0.1; a =1 m/s/s, right )

Since there is no vertical acceleration, the normal force is equal to the gravity force.

The mass can be found using the equation F_grav = m * g.

Using "mu" = F_frict / F_norm, "mu" = (10 N) / (100 N) = 0.1.

The F_net is the vector sum of all the forces: 100 N, up plus 100 N, down equals 0 N. And 20 N, right plus 10 N, left = 10 N, right.

Finally, a = F_net / m = (10 N) / (10.2 kg) = 0.980 m/s/s.

Practice #3

C onsider the two-body situation at the right. A 100.0-gram hanging mass (m2) is attached to a 325.0-gram mass (m1) at rest on the table. The coefficient of friction between the 325.0-gram mass and the table is 0.215. Determine the acceleration of the system and the tension in the string.

 a = 0.695 m/s²  F_tens = 0.911 N

The solution here will use the approach of a free-body diagram and Newton's second law analysis of each indivi-dual mass. The free-body diagrams for the two objects are shown below.

For mass m₁: F_grav = m₁•g = (0.3250 kg) • (9.8 N/kg) = 3.185 N ,

F_norm = F_grav = 3.185 N F_frict = µ•F_norm = (0.215)•(3.185 N) = 0.68478 N

Applying Newton's second law to m₁:

Equation 1: F_tens - 0.68478 = 0.3250•a

For mass m₂: F_grav = m₂•g = (0.1000 kg)•(9.8 N/kg) = 0.9800 N

Applying Newton's second law to m₂:

Equation 2: 0.9800 - F_tens = 0.1000•a

Combining Equations 1 and 2 leads to the answers:

a = 0.69464 m/s² = ~0.695 m/s² F_tens = 0.91054 N = ~0.911 N