87.1k views
5 votes
A box with a mass of 50 kg is at rest on a floor. The box is pushed 4.0 m along the floor with one horizontal force of 110 N. Then the box slides 1.1 m before stopping.

a) What is the coefficient of friction?
b) What is the maximum speed of the box?

User Notacorn
by
8.0k points

1 Answer

3 votes

The coefficient of friction can be determined using the formula:

Coefficient of friction = Force of friction / Normal force

To find the force of friction, we can use Newton's second law:

Force of friction = Mass × Acceleration

Since the box is at rest, the acceleration is 0. Therefore, the force of friction is also 0.

The normal force is equal to the weight of the box, which can be calculated using the formula:

Weight = Mass × Gravitational acceleration

Substituting the given values:

Weight = 50 kg × 9.8 m/s^2 = 490 N

Now we can calculate the coefficient of friction:

Coefficient of friction = 0 / 490 N = 0

Therefore, the coefficient of friction is 0.

To find the maximum speed of the box, we need to calculate the acceleration first. We can use Newton's second law again:

Force = Mass × Acceleration

The net force acting on the box is the horizontal force minus the force of friction:

Net force = 110 N - 0 N = 110 N

Substituting the values:

110 N = 50 kg × Acceleration

Solving for acceleration:

Acceleration = 110 N / 50 kg = 2.2 m/s^2

Now, we can use the equation of motion to find the maximum speed:

v^2 = u^2 + 2as

Where:
v = final velocity (maximum speed)
u = initial velocity (0 m/s)
a = acceleration (2.2 m/s^2)
s = distance (1.1 m)

Substituting the values:

v^2 = 0^2 + 2 × 2.2 m/s^2 × 1.1 m
v^2 = 4.84 m^2/s^2
v = √4.84 m^2/s^2
v ≈ 2.2 m/s

Therefore, the maximum speed of the box is approximately 2.2 m/s.

User Benedict Lee
by
8.5k points

No related questions found