87.9k views
2 votes
A 4 kg box is at rest on a table. The coefficient of friction are 0.30 and 0.10 for static and kinetic respectively. Then a 10N horizontal force is applied to box.

a. What is the Normal Force acting on the box?
b. What is the value of the Friction Force?
c. What is the Net Force?
d. What is the acceleration of the box?

User Magdaline
by
7.6k points

1 Answer

4 votes

Answer:

Step-by-step explanation:

a. normal force is mass * acceleration due to gravity, so the normal force is = 4 * 9.8, which is 39.2 N.

b. friction force is normal force * the coefficient of the friction, for static friction: .3 * 39.2 = 11.76 N. For kinetic friction: .1 * 39.2 = 3.92 N.

c. net force is the sum of all force apply on the object, since the object is rest at a table, then the only force apply on the object is the horizontal force, which is 10N, since the static force is greater than the force apply, then our net force is -1.76N( 10 - 11.76).

d. the acceleration of the box is 0 because the box is not moving.

a : normal force = 39.2N

b : static force = 11.76N, kinetic force = 3.92N

c : net force = -1.76N

d : acceleration = 0

User Edwin Cheong
by
7.6k points

No related questions found