A 50-kg block is pushed a distance of 5.0 m across a floor by a horizontal force Fp whose magnitude is 150 N. Fp is parallel to the displacement of the block. The coefficient of kinetic friction is 0.25.

A) What is the total work done on the block?
B) If the box started from rest, what is the final speed of the block?

weight = 50g = 490 N

friction force = .25 * 490 = 122.5 N
net force = 150 - 122.5 = 27.5 N
force * distance = 27.5*5 = 137.5 Joules work input

.5 m v^2 = 137.5
v^2 = 5.5
v =2.35 at the end

A) Well, let's calculate the total work done on the block. The work done is equal to the force multiplied by the distance moved in the direction of the force. Since the force is parallel to the displacement, we can simply use the equation:

Work = Force * Distance

So, the work done on the block is:

Work = 150 N * 5.0 m
Work = 750 J

B) Now, let's calculate the final speed of the block. To do this, we need to take into account the work done and the friction involved. The work done is equal to the change in kinetic energy of the block, so:

Work = ΔK
ΔK = 750 J

The initial kinetic energy of the block is zero since it starts from rest. The final kinetic energy can be calculated using the equation:

K = (1/2)mv²

So, the final kinetic energy of the block is:

ΔK = (1/2)mv² - (1/2)mv₀²

Since the block starts from rest, the initial velocity (v₀) is zero, so the equation simplifies to:

ΔK = (1/2)mv² - 0

Plugging in the given values, we have:

750 J = (1/2)(50 kg)v²
15 = v²

Taking the square root, we get:

v = √15 m/s

So, the final speed of the block is approximately 3.87 m/s.

A) The total work done on the block can be calculated using the equation:

Work = Force × Distance × Cos(θ)

Where:
- Work is the total work done on the block
- Force is the parallel component of the applied force
- Distance is the displacement of the block
- θ is the angle between the applied force and the displacement

In this case, the force Fp is parallel to the displacement of the block, so the angle θ is 0 degrees. Therefore, the equation simplifies to:

Work = Force × Distance

Plugging in the values, we have:

Work = 150 N × 5.0 m = 750 J

Therefore, the total work done on the block is 750 Joules.

B) To calculate the final speed of the block, we can use the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy.

The work done on the block is equal to the force of friction multiplied by the distance:

Work = Force of Friction × Distance

The force of friction can be calculated using the equation:

Force of Friction = Coefficient of Friction × Normal Force

The normal force is equal to the weight of the block, which is given by the equation:

Normal Force = Mass × gravitational acceleration

Plugging in the values, we have:

Force of Friction = 0.25 × (50 kg × 9.8 m/s^2) = 122.5 N

Now we can calculate the work done:

Work = 122.5 N × 5.0 m = 612.5 J

Since the work done is negative (opposite direction of motion), we can use the equation:

Work = ΔKE

where ΔKE is the change in kinetic energy. The initial kinetic energy is zero since the block starts from rest. Therefore:

-612.5 J = ΔKE

Solving for ΔKE, we have:

ΔKE = -612.5 J

Since the initial kinetic energy is zero, the final kinetic energy is equal to ΔKE:

KE = -612.5 J

The final speed of the block can be calculated using the equation for kinetic energy:

KE = 0.5 × mass × velocity^2

Plugging in the values, we have:

-612.5 J = 0.5 × 50 kg × velocity^2

Solving for velocity, we have:

velocity^2 = (-612.5 J) / (0.5 × 50 kg) = -24.5 m^2/s^2

Since velocity cannot be negative, there is an error in the calculation. Please review the numbers and calculations provided to determine the correct final speed of the block.

To solve this problem, we need to consider various factors such as the work done by the applied force, the work done by friction, and the work-energy theorem.

A) To find the total work done on the block, we need to calculate the work done by the applied force and the work done by friction separately.

Work done by the applied force:
The work done by a force is given by the formula: work = force * displacement * cos(theta), where theta is the angle between the force and the displacement.

In this case, the applied force (Fp) is parallel to the displacement of the block, so the angle between them is 0 degrees. Therefore, the work done by the applied force is:
work_applied = Fp * displacement * cos(0) = Fp * displacement

Substituting the known values:
work_applied = 150 N * 5.0 m = 750 J

Work done by friction:
The work done by friction can be calculated using the formula: work_friction = force_friction * displacement * cos(180), where force_friction = coefficient of friction * normal force.

The normal force can be found using the formula: normal force = mass * acceleration due to gravity.

normal force = 50 kg * 9.8 m/s^2 = 490 N

force_friction = coefficient of kinetic friction * normal force = 0.25 * 490 N = 122.5 N

Since the force of friction opposes the displacement, the angle between them is 180 degrees.

work_friction = force_friction * displacement * cos(180) = force_friction * displacement * (-1)

Substituting the known values:
work_friction = 122.5 N * 5.0 m * (-1) = -612.5 J

Total work done on the block:
The total work done on the block is the sum of the work done by the applied force and the work done by friction.

total work = work_applied + work_friction
total work = 750 J + (-612.5 J) = 137.5 J

Therefore, the total work done on the block is 137.5 Joules.

B) The final speed of the block can be calculated using the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy.

The initial kinetic energy of the block is zero since it starts from rest. Therefore, the work done on the block is equal to its final kinetic energy.

Using the formula: work = (1/2) * mass * velocity^2

Substituting the known values:
137.5 J = (1/2) * 50 kg * final velocity^2

Simplifying the equation:
275 = 25 * final velocity^2

Dividing by 25:
final velocity^2 = 11

Taking the square root of both sides:
final velocity = √11 m/s

Therefore, the final speed of the block is approximately 3.32 m/s.