A 220 N crate is pushed along the ground with a horizontal force. The coefficient of kinetic friction is 0.21 and the coefficient of static friction is 0.31. The crate moves at a constant speed of 4.1 m/s.

a) Draw a free-body diagram for the crate.

b) What is the size of the pushing force? 46.2 N

c) How long will it take the box to stop if the pushing force is removed?

Fc = 220N @ 0deg. = Force of crate.

Fp = 220sin(0) = 0 = Force parallel to
ground.
Fv = 220cos(0) = 220N. = Force perpendicular to ground.

Fs = u*Fv = 0.31 * 220 = 68.2N. = Force
of static friction.
Fk = 0.21 * 220 = 46.2N. = Force of kinetic friction.

b. Fn = Fap - Fp - Fk = ma = 0, a = 0.
Fap - 0 - 46.2 = 0,
Fap=46.2N.=Force applied(pushing force)

a) To draw a free-body diagram for the crate, we need to consider all the forces acting on it. The main forces are:

1) Weight (W): This force acts vertically downwards and can be calculated using the formula W = mg, where m is the mass of the crate and g is the acceleration due to gravity.

2) Normal force (N): This force acts perpendicular to the surface and balances the weight of the crate. The magnitude of the normal force is equal to the weight of the crate when it is on a horizontal surface.

3) Friction force (F): There are two types of friction forces to consider - kinetic friction and static friction. The kinetic friction force acts parallel to the surface and its magnitude can be calculated using the formula F = μkN, where μk is the coefficient of kinetic friction and N is the normal force. The static friction force acts parallel to the surface and its magnitude can be calculated using the formula F = μsN, where μs is the coefficient of static friction and N is the normal force.

b) To find the size of the pushing force, we need to create a force balance equation. Since the crate is moving at a constant speed, the net force acting on it is zero. Therefore, the pushing force is equal in magnitude but opposite in direction to the kinetic friction force. Mathematically, we can write:

Pushing force = Friction force
Pushing force = μkN

Given that the coefficient of kinetic friction is 0.21, we can substitute this value into the equation. However, we need to first calculate the normal force.

To calculate the normal force, we use the equation N = mg, where m is the mass of the crate. The weight of the crate can be calculated using the formula W = mg, which implies that the weight is equal to the normal force (since the crate is on a horizontal surface and not accelerating vertically).

c) If the pushing force is removed, the only force acting on the crate would be the kinetic friction force, which acts against the motion. The crate will decelerate until the net force acting on it is zero. This means that the kinetic friction force will cause the crate to stop.

To calculate the time it takes for the box to stop, we need to use the equation of motion:

v = u + at

Where:
v = final velocity (0 m/s, since the box stops)
u = initial velocity (4.1 m/s, given in the question)
a = acceleration (negative, as the box is decelerating)
t = time

Since the box is moving in the opposite direction of its initial velocity, the acceleration is given by:

a = -F/m

Where F is the friction force and m is the mass of the crate.

Finally, we can substitute the values into the equation and solve for t.