A load of mass 100 kg is pulled by a horse on a horizontal ground with a constant horizontal force 1000 N. The frictional force between the load and the horizontal ground is 200 N. What is the velocity of the load and the work done by the horse after the load is pulled through a distance of 4.0 m?
M = 100kg
F = 1800N
S = 4.0m
W = F S
W = (1800N)(4.0m) = 7200Nm
W = 7.2 kJ
A = F / M
V = V0 + A T
S = S0 + V0 T + ½ A T^2
Assume initial velocity and displacement are 0.
S = ½ F T^2 / m
T = √(2 M S / F)
V = F √(2 M S / F) / M
V = √(2 F S / M)
V = √(2 1800N 4.0m / 100kg)
V = 12 m/s
To find the velocity of the load, we need to calculate the acceleration of the load first.
The net force acting on the load can be found by subtracting the frictional force from the applied force:
Net Force = Applied Force - Frictional Force
= 1000 N - 200 N
= 800 N
We can use Newton's second law, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration:
Net Force = Mass * Acceleration
Rearranging the equation to find the acceleration:
Acceleration = Net Force / Mass
= 800 N / 100 kg
= 8 m/s^2
Next, we can use the kinematic equation to find the final velocity of the load:
v^2 = u^2 + 2as
Here, u is the initial velocity (assumed to be 0 m/s), a is the acceleration, and s is the distance traveled.
Plugging in the values:
v^2 = 0^2 + 2 * 8 m/s^2 * 4 m
v^2 = 64 m^2/s^2
Taking the square root of both sides:
v = √(64 m^2/s^2)
v = 8 m/s
Therefore, the velocity of the load is 8 m/s.
To calculate the work done by the horse, we can use the equation:
Work = Force * Distance * cos(θ)
Here, Force is the applied force, Distance is the distance traveled, and θ is the angle between the force and the displacement (which is 0 degrees since it is a horizontal force).
Plugging in the values:
Work = 1000 N * 4.0 m * cos(0)
Work = 4000 N * 1
Work = 4000 J
Therefore, the work done by the horse after the load is pulled through a distance of 4.0 m is 4000 Joules.
To find the velocity of the load, we can use Newton's second law of motion, which states that the net force acting on an object is equal to its mass multiplied by its acceleration. In this case, the net force is the force applied by the horse minus the frictional force.
Net force = Applied force - Frictional force
= 1000 N - 200 N
= 800 N
Now, we can find the acceleration of the load using the formula:
Net force = mass × acceleration
Rearranging the formula, we have:
Acceleration = Net force / mass
= 800 N / 100 kg
= 8 m/s²
Since the acceleration is constant, we can use the following kinematic equation to find the final velocity:
vf² = vi² + 2ad
where vf is the final velocity, vi is the initial velocity (which we assume to be 0, in this case), a is the acceleration, and d is the displacement.
Plugging in the values, we have:
vf² = 0 + 2 × 8 m/s² × 4.0 m
vf² = 64 m²/s²
Taking the square root of both sides:
vf = √64 m²/s²
vf = 8 m/s
Therefore, the velocity of the load is 8 m/s.
To find the work done by the horse, we can use the formula:
Work = Force × Distance
The force applied by the horse is the same as the force used to calculate the acceleration, which is 800 N. The distance traveled by the load is given as 4.0 m.
Plugging in the values, we have:
Work = 800 N × 4.0 m
Work = 3200 Nm
Therefore, the work done by the horse is 3200 Nm or 3200 Joules.