A 0.500 kg object is thrown vertically upward with an average applied force of 8.20 N by a student. The force is applied through a displacement of 1.50m.
a. What is the average net force acting on the object?
b. What is the velocity of the object when it leaves the student's hand? (Assume initial velocity is zero)
8.2 UP and .5(9.81 ) down
= 3.295 Newtons up
Work done
=Force * 1.5 meters = 3.295*1.5
= 4.9425 Joules
so
(1/2) (.5) v^2 = 4.9425
v = 4.45 m/s
a. Well, if the average applied force is 8.20 N and the displacement is 1.50 m, we can calculate the average net force by using the formula: average net force = applied force / displacement. So, the average net force would be 8.20 N / 1.50 m = 5.46 N.
b. Now, to find the velocity of the object when it leaves the student's hand, we can use the equation: final velocity = initial velocity + (acceleration x time). Since the initial velocity is zero, we only need to find the acceleration and time.
The acceleration can be calculated using Newton's second law: acceleration = net force / mass. In this case, the net force is 5.46 N and the mass is 0.500 kg. So, the acceleration is 5.46 N / 0.500 kg = 10.92 m/s^2.
Now let's find the time it takes for the object to reach its final velocity. We can use the formula: final velocity = initial velocity + (acceleration x time). Since the initial velocity is zero and we want to find the time, we can rearrange the equation to: time = (final velocity - initial velocity) / acceleration. Plugging in the values, we get: time = (0 - 0) / 10.92 m/s^2 = 0 seconds.
Since the time is zero, that means the object leaves the student's hand instantly. So the velocity of the object when it leaves the student's hand is also zero. But hey, don't worry, it'll pick up speed soon enough!
a. To find the average net force acting on the object, we can use Newton's second law of motion:
F_net = m * a
where F_net is the net force, m is the mass of the object, and a is the acceleration.
In this case, the mass of the object is 0.500 kg. The acceleration can be calculated using the work-energy theorem:
W = F * d * cos(theta)
where W is the work done on the object, F is the applied force, d is the displacement, and theta is the angle between F and d.
Since the force is applied vertically upward and the displacement is also vertical, the angle between them is 0 degrees, so cos(theta) = 1.
Plugging in the given values:
W = F * d * cos(theta)
W = 8.20 N * 1.50 m * 1
W = 12.3 J
The work done on the object is equal to the change in kinetic energy, so:
W = Delta KE
If the object starts from rest, the initial kinetic energy is zero, so:
12.3 J = Delta KE
Delta KE = 12.3 J
The change in kinetic energy can be calculated using the equation:
Delta KE = (1/2) * m * (v_f^2 - v_i^2)
where v_f is the final velocity and v_i is the initial velocity.
Since the initial velocity is zero, the equation simplifies to:
12.3 J = (1/2) * 0.500 kg * v_f^2
Solving for v_f:
v_f^2 = (2 * 12.3 J) / (0.500 kg)
v_f^2 = 49.2 m^2/s^2
v_f = sqrt(49.2) m/s
v_f = 7.01 m/s
b. The velocity of the object when it leaves the student's hand is 7.01 m/s.
To find the average net force acting on the object, you can use the formula:
Net force = mass × acceleration
First, let's determine the acceleration of the object. We can use Newton's second law of motion:
F = m × a
Where:
F is the applied force, which is 8.20 N.
m is the mass of the object, which is 0.500 kg.
a is the acceleration.
Rearranging the equation, we can solve for acceleration:
a = F / m
Substituting the given values:
a = 8.20 N / 0.500 kg
a = 16.4 m/s²
Now that we have the acceleration, we can find the net force:
Net force = (mass) × (acceleration)
Net force = 0.500 kg × 16.4 m/s²
Net force = 8.20 N
Therefore, the average net force acting on the object is 8.20 N.
To find the velocity of the object when it leaves the student's hand, we can use the kinematic equation:
vf = vi + at
Where:
vf is the final velocity,
vi is the initial velocity (which is 0 m/s),
a is the acceleration (which we have already calculated),
and t is the time.
To find the time, we need to know the total displacement of the object. In this case, it is given as 1.50 m. Since the object is thrown vertically upward, we know that the displacement is positive.
The equation for displacement in terms of initial velocity, acceleration, and time is:
δx = vi × t + (1/2)at²
Substituting the values, we get:
1.50 m = 0 m/s × t + (1/2)(16.4 m/s²)t²
Simplifying the equation:
1.50 m = (1/2)(16.4 m/s²)t²
Multiplying both sides by 2 to remove the fraction:
3.0 m = (16.4 m/s²)t²
Dividing both sides by 16.4 m/s²:
t² = 3.0 m / 16.4 m/s²
t² = 0.183 m²/s²
Taking the square root of both sides:
t ≈ √0.183 m²/s²
t ≈ 0.428 s
Now that we have the time, we can calculate the final velocity:
vf = vi + at
vf = 0 m/s + (16.4 m/s²)(0.428 s)
vf ≈ 7.01 m/s
Therefore, the velocity of the object when it leaves the student's hand is approximately 7.01 m/s.