A 1.0 kg object is given a push along a horizontal surface. If the velocity of the object when it is released is 0.50 m/s west, and the object slides 0.25m before coming to a stop, what is the magnitude of the force of friction?

.25 = 1/2 at^2

v = .5 - at
F = 1.0 a

Well, let me put on my detective hat and solve this mystery for you. We know that the object comes to a stop after sliding 0.25 m. This means that the net force acting on the object must be in the opposite direction of its initial velocity.

Since we're dealing with a horizontal surface, the only force opposing the motion of the object is the force of friction. And according to Newton's second law, the force of friction is equal to the mass of the object multiplied by its acceleration.

Now, we need to find the acceleration of the object. We can use the equation of motion:

v^2 = u^2 + 2as

where v is the final velocity (zero in this case), u is the initial velocity (0.50 m/s west), a is the acceleration, and s is the distance traveled (0.25 m).

Rearranging the equation, we have:

a = (v^2 - u^2) / (2s)

a = (0 - (0.50)^2) / (2 * 0.25)

a = -0.25 / 0.5

a = -0.5 m/s^2

Since the acceleration is negative, this means the force of friction is in the East direction to oppose the object's Westward motion.

Now, we can calculate the magnitude of the force of friction by using Newton's second law:

Force of friction = mass * acceleration

Force of friction = 1.0 kg * 0.5 m/s^2

Force of friction = 0.5 N

So, the magnitude of the force of friction is 0.5 Newtons.

Now that the mystery is solved, I'll be taking off my detective hat and putting on my clown nose! 🤡

To find the magnitude of the force of friction, we can use the concept of work and energy. The work done by the force of friction is equal to the change in kinetic energy of the object.

1. First, let's find the initial kinetic energy (K₁) of the object. The formula for kinetic energy is given by K = (1/2)mv², where m is the mass of the object and v is the velocity.

The mass (m) of the object is 1.0 kg, and the velocity (v) is 0.50 m/s. Therefore, the initial kinetic energy is:
K₁ = (1/2)(1.0 kg)(0.50 m/s)² = 0.125 J

2. Next, let's find the final kinetic energy (K₂) of the object. The final kinetic energy is zero since the object comes to a stop.

K₂ = 0 J

3. The change in kinetic energy (ΔK) is given by ΔK = K₂ - K₁.

ΔK = 0 J - 0.125 J = -0.125 J

4. The work done by the force of friction (W) is equal to the negative change in kinetic energy, W = -ΔK.

W = -(-0.125 J) = 0.125 J

5. The work done by the force of friction is equal to the force of friction (F) multiplied by the displacement (d) over which it acts.

W = F * d

Since the object slides 0.25 m before coming to a stop, we have:

0.125 J = F * 0.25 m

6. Now, let's solve for the magnitude of the force of friction (F).

F = 0.125 J / 0.25 m
F = 0.5 N

Therefore, the magnitude of the force of friction acting on the object is 0.5 N.

To find the magnitude of the force of friction, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration. In this case, we know the mass of the object (1.0 kg) and we can calculate the acceleration by using the kinematic equation:

v² = u² + 2as

where:
v = final velocity (0 m/s, since the object comes to a stop)
u = initial velocity (0.50 m/s west)
a = acceleration
s = distance traveled (0.25 m)

Rearranging the equation to solve for acceleration, we get:

a = (v² - u²) / (2s)

Substituting the known values, we get:

a = (0 - (0.50)²) / (2 * 0.25)

Simplifying the expression inside the parentheses first:

a = (0 - 0.25) / 0.5

a = -0.5 m/s²

The negative sign indicates that the acceleration is in the opposite direction of the initial velocity, which makes sense since the object is slowing down.

Now, we can use Newton's second law to find the net force acting on the object:

F_net = m * a

Substituting the values we know:

F_net = 1.0 kg * (-0.5 m/s²)

F_net = -0.5 N

The net force is negative because it acts in the opposite direction of the initial velocity of the object.

Since the only force acting on the object in the horizontal direction is the force of friction, the magnitude of the force of friction is equal to the magnitude of the net force, which is 0.5 N.

The answer would be 279 N because when you rearrange the formula F(net)=Fa-Fg and plug in your numbers for Fnet=(132N) in addition to your force of gravity of 147.15N you will have an answer of 279.