A 35.2 kg wagon is towed up a hill inclined at 18.3◦ with respect to the horizontal. The tow rope is parallel to the incline and exerts a force of 125 N on the wagon. Assume that the wagon starts from rest at the bottom of the hill, and disregard friction.
The acceleration of gravity is 9.81 m/s2 .
How fast is the wagon going after moving 75.4 m up the hill?
Answer in units of m/s
Fw = m*g = 35.2 * 9.8 = 345 N. = Force of the wagon.
Fp = 345*sin18.3 = 108.3 N. = Force parallel to the hill.
a = (Fex-Fp)/m = (125-108)/35.2 = 0.474
m/s^2.
V^2 = Vo^2 + 2a*d = 0 + 0.949*75.4 = 71.54
V = 8.46 m/s.
Well, well, well, we've got ourselves a little physics problem here! I'm going to calculate the wagon's speed using a little bit of clown-venient mathematics.
The force exerted by the tow rope is parallel to the incline of the hill, so we can decompose it into two components: one parallel to the incline and one perpendicular to the incline.
The component parallel to the incline will be the one that affects the motion of the wagon. To calculate it, we use the formula:
Force_parallel = Force_total * sin(angle)
= 125 N * sin(18.3°)
Now, we can use Newton's second law, F = ma, to find the acceleration of the wagon:
Force_parallel = mass * acceleration
125 N * sin(18.3°) = 35.2 kg * acceleration
Solving for acceleration gives us:
acceleration = (125 N * sin(18.3°)) / 35.2 kg
Now, we can use the kinematic equation, v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity (zero in this case), a is the acceleration, and s is the displacement.
Plugging in the values, we get:
v^2 = 0^2 + 2 * (acceleration) * (displacement)
v^2 = 2 * [(125 N * sin(18.3°)) / 35.2 kg] * 75.4 m
Now, we just need to calculate the square root of that mess to find the wagon's speed:
v = √[2 * (125 N * sin(18.3°)) / 35.2 kg * 75.4 m]
After playing around with the numbers, the wagon's speed comes out to be approximately 4.27 m/s.
So, after all that math, the wagon is moving at a speed of 4.27 m/s. Now that's what I call wag-on and wag-off the hill!
To find the speed of the wagon after moving up the hill, we can use the work-energy principle. The work done by the force applied by the tow rope will be equal to the change in kinetic energy of the wagon.
The work done by the force applied by the tow rope can be calculated as:
Work = force * distance * cos(angle)
where,
force = 125 N
distance = 75.4 m
angle = 18.3°
Work = 125 N * 75.4 m * cos(18.3°)
Next, we need to find the change in kinetic energy of the wagon. At the bottom of the hill, the wagon is at rest, so its initial kinetic energy is zero. The final kinetic energy of the wagon after moving up the hill will be:
Final kinetic energy = 1/2 * mass * (speed)^2
where,
mass = 35.2 kg
speed = final velocity
Since the work done is equal to the change in kinetic energy, we can equate the two quantities:
Work = Final kinetic energy - Initial kinetic energy
125 N * 75.4 m * cos(18.3°) = 1/2 * 35.2 kg * speed^2 - 0
Simplifying the equation:
9447 N·m = 1/2 * 35.2 kg * speed^2
speed^2 = (9447 N·m * 2) / (35.2 kg)
speed^2 = 535.65 m^2/s^2
Taking the square root of both sides:
speed = sqrt(535.65 m^2/s^2)
speed ≈ 23.15 m/s
Therefore, the speed of the wagon after moving 75.4 m up the hill is approximately 23.15 m/s.
To find the speed of the wagon after moving up the hill, we can use the principles of Newton's second law and the work-energy theorem.
First, let's find the net force acting on the wagon. Since the wagon is being towed parallel to the incline, the force components acting on the wagon are the gravitational force acting downward (mg) and the force of the tow rope acting upward (F).
The gravitational force, mg, can be calculated as:
mg = mass * acceleration due to gravity
mg = 35.2 kg * 9.81 m/s^2
Next, let's resolve the force of the tow rope (F) into its components parallel and perpendicular to the incline. Since the tow rope is parallel to the incline, the perpendicular component does not affect the motion of the wagon up the hill.
The force component acting parallel to the incline can be found using trigonometry:
F_parallel = F * sin(18.3°)
Now, let's calculate the net force acting on the wagon:
net force = F_parallel - mg
Next, we can use Newton's second law to find the acceleration of the wagon:
net force = mass * acceleration
acceleration = net force / mass
With the acceleration known, we can use the kinematic equation to find the final velocity of the wagon after moving 75.4 m up the hill. Since the wagon starts from rest and moves only along the incline, we can use the following equation:
v^2 = u^2 + 2as
where:
v = final velocity
u = initial velocity (0 m/s since the wagon starts from rest)
a = acceleration
s = distance moved up the incline (75.4 m)
Solving for v, we get:
v = √(2as)
Now, we can substitute the values we know into the equation and calculate the final velocity:
v = √(2 * acceleration * distance)
Remember to convert the angle to radians when taking the sine of it.
Finally, just plug in the values and calculate the final velocity.