A stock room worker pushes a box with mass 11.2 kg on a horizontal surface with a constant speed of 4.50 m/s. The coefficient of kinetic friction between the box and the surface is 0.25. a) What horizontal force must be applied by the worker to maintain the motion? b) If the force calculated in parts (a) is removed, how far does the box slide before coming to rest
a = 0
so the sum of forces is 0
F is applied
- mu m g is friction force
so
F = mu m g
= .25 * 11.2 * 9.81
then
m a = mu m g
a = mu g
solve for a
time to stop first
0 = 4.50 - a t
solve for t
then
x = Xo + Vi t + (1/2) a t^2
x = 0 + 4.5 t - .5 a t^2
Vfinal = vi + at not negative it should be + sign i don't know pla reply
a) To maintain the motion of the box at a constant speed, the worker must apply a force equal to the force of kinetic friction. The formula for the force of kinetic friction is given by:
fk = μk * N
where fk is the force of kinetic friction, μk is the coefficient of kinetic friction, and N is the normal force.
The normal force is equal to the weight of the box, which is given by:
N = m * g
where m is the mass of the box and g is the acceleration due to gravity.
N = 11.2 kg * 9.8 m/s^2 = 109.76 N
Now we can calculate the force of kinetic friction:
fk = 0.25 * 109.76 N = 27.44 N
Therefore, the horizontal force that must be applied by the worker to maintain the motion is 27.44 N.
b) If the applied force is removed, the box will decelerate due to the force of kinetic friction until it comes to rest. The deceleration can be determined using Newton's second law:
fk = m * a
where fk is the force of kinetic friction, m is the mass of the box, and a is the acceleration.
From part a), we know that fk = 27.44 N. So we can rearrange the equation to solve for acceleration:
a = fk / m = 27.44 N / 11.2 kg = 2.45 m/s^2
Now we can use the following equation to find the distance traveled by the box before coming to rest:
v^2 = u^2 + 2as
where v is the final velocity (0 m/s), u is the initial velocity (4.50 m/s), a is the acceleration (-2.45 m/s^2), and s is the distance traveled.
0^2 = 4.50^2 + 2(-2.45)s
Solving for s:
0 = 20.25 - 4.9s
4.9s = 20.25
s = 20.25 / 4.9
s ≈ 4.14 m
Therefore, the box will slide approximately 4.14 meters before coming to rest.
To find the horizontal force that the worker must apply to maintain the constant motion of the box, we need to consider the forces acting on the box.
a) The forces involved are the force that the worker applies (F_applied), the force of gravity (F_gravity), and the frictional force (F_friction). Since the box is moving at a constant speed, the net force acting on it must be zero.
The force of gravity (F_gravity) acting on the box can be calculated by multiplying the mass of the box (m) by the acceleration due to gravity (g).
F_gravity = m * g
where m = 11.2 kg (mass of the box) and g = 9.8 m/s² (acceleration due to gravity).
F_gravity = 11.2 kg * 9.8 m/s² = 109.76 N
The frictional force (F_friction) can be calculated using the formula:
F_friction = coefficient of friction * Normal force
where the coefficient of kinetic friction (μ) is given as 0.25.
The normal force (F_normal) is equal to the force of gravity acting on the box (F_gravity).
F_normal = F_gravity = 109.76 N
F_friction = 0.25 * 109.76 N = 27.44 N
Since the net force is zero, the force applied by the worker (F_applied) must be equal in magnitude but opposite in direction to the frictional force (F_friction).
F_applied = -F_friction = -27.44 N
Therefore, the horizontal force that the worker must apply to maintain the motion of the box is 27.44 N (opposite in direction to the frictional force).
b) If the force calculated in part (a) is removed, the box will decelerate due to the kinetic friction acting in the opposite direction. To find out how far the box slides before coming to rest, we can use the equation of motion:
v² = u² + 2as
where v = 0 m/s (final velocity), u = 4.50 m/s (initial velocity), a = acceleration, and s = distance.
Since the box comes to rest, the final velocity (v) is 0 m/s.
0² = (4.50 m/s)² + 2 * a * s
Rearranging the equation, we get:
a * s = - (4.50 m/s)²
We know that the acceleration (a) is equal to the frictional force (F_friction) divided by the mass of the box (m).
a = F_friction / m = 27.44 N / 11.2 kg ≈ 2.45 m/s²
Plugging in the values:
2.45 m/s² * s = - (4.50 m/s)²
s = - (4.50 m/s)² / (2.45 m/s²)
s ≈ -9.83 m²/s² / 2.45 m/s²
s ≈ -4.01 m
Since distance cannot be negative, the box slides for approximately 4.01 m before coming to rest.