Question

A block of mass m lies on a horizontal table. The coefficient of static friction between the block and the table is ?s. The coefficient of kinetic friction is ?k, with ?k<?s.

1)If the block is at rest (and the only forces acting on the block are the force due to gravity and the normal force from the table), what is the magnitude of the force due to friction?

2)Suppose you want to move the block, but you want to push it with the least force possible to get it moving. With what force

F must you be pushing the block just before the block begins to move?

Express the magnitude of F in terms of some or all the variables ?s, ?k, and m, as well as the acceleration due to gravityg.

3)Suppose you push horizontally with half the force needed to just make the block move. What is the magnitude of the friction force?

Express your answer in terms of some or all of the variables ?s, ?k, and m, as well as the acceleration due to gravity g.

4)Suppose you push horizontally with precisely enough force to make the block start to move, and you continue to apply the same amount of force even after it starts moving. Find the acceleration

a of the block after it begins to move.

Express your answer in terms of some or all of the variables ?s, ?k, and m, as well as the acceleration due to gravity g.

Answer 1

1. Zero. Because there is no net motion.

2. F= ?smg is the minimum required force where m is the mass of the block and g is acceleration due to gravity.

3. Since the block is not moving there opposing frictional force must be equal to the force applied. Therefore friction is equal to the applied force = (?smg)/2

4. Once the block starts moving the frictional coefficient is ?k. Net force acting is ?smg-?kmg. Therefore acceleration is force by mass = (?s-?k)g.

Answer 2

1.
The friction force is always opposite in direction to the motion along the surface. No motion means no friction.
F_friction = 0.

2.
When the block is on the point of moving:
F_friction = mu_s mg

As the acceleration is 0:
F - F_friction = 0

Therefore F = mu_s mg.

3.
F = mu_s mg / 2
F - F-friction = 0

Therefore:
F_friction = mu_s mg / 2.

4.
When the block is on the point of moving:
F = mu_s mg

When the block is moving with acceleration a:
F - mu_k mg = ma

Eliminating F:
(mu_s - mu_k)mg = ma
a = (mu_s - mu_k)g.

Answer 3

1) zero.

2)F=mu(s)*mg [mu(s) is static friction coefficient].

3)F/2 = mu(s)*mg/2.

4)a= (F-f)/m

=[mu(s)*mg - mu(k)*mg]/m

=[mu(s)- mu(k)]*g