1. A monkey in a perch 20 m high in a tree drops a coconut above the head of a zoo keeper as he runs with a speed 1.5 m/s beneath the tree actually intending to hit the toes of the zoo keeper, how early in seconds should the coconut be dropped by the monkey.
2 A monkey in a perch 20 m high in a tree drops a coconut above the head of a zoo keeper as he runs with a speed 1.5 m/s beneath the tree. How far behind him in meters does the coconut hit the ground.
3. A man leaves the garrage in the house and drive to a neighboring town which is twenty km away from his house on sight-seeing. He return home to his garrage two hours after. What is his average velocity from his home in km/h?
4. Two force act a point object as follows: 100 N at 170 degree and 100 N at 50 degree. Find the resultant force.
5. A 40 N force applied at an angle of 37 degree above the horizontal pulls a 5-kg box on a horizontal floor. The acceleration of the box is 3m/s square. How large a frictional force must be retarding the motion of the box?
3. d = V*T = 40 km
V*2 = 40
V = 20 km/h
4. Fr = 100N[170o] + 100N[50o]
X = 100*Cos170 + 100*Cos50 = -34.2 N.
Y = 100*sin170 + 100*sin50 = 94 N.
Q2.
Tan A = Y/X = 94/-34.2 = -2.74854
A = -70o = 70o N. of W. = 110o CCW
Fr = Y/sin A = 94/sin110 =
5. Fap*CosA-Fk = M*a
Fap = 40 N.
A = 37o
Fk = Force of kinetic friction.
M = 5 kg
a = 3 m/s^2
Solve for Fk
1. To determine how early the coconut should be dropped, we need to consider the time it takes for the coconut to fall and the time it takes for the zookeeper to reach a position where the coconut should hit his toes.
First, let's find the time it takes for the coconut to fall to the ground. We can use the equation for free fall:
h = (1/2)gt^2
where h is the height (20 m) and g is the acceleration due to gravity (approximately 9.8 m/s^2). Rearranging the equation to solve for time (t), we get:
t = sqrt(2h/g)
Plugging in the values, we have:
t = sqrt(2 * 20 / 9.8) = 2.02 seconds
Now, let's find the distance the zookeeper will cover in that time. The distance can be calculated using the equation:
d = v * t
where d is the distance, v is the velocity (1.5 m/s), and t is the time (2.02 seconds). Plugging in the values, we have:
d = 1.5 * 2.02 = 3.03 meters
Therefore, the monkey should drop the coconut 3.03 meters in advance to hit the zookeeper's toes.
2. To find how far behind the zookeeper the coconut hits the ground, we need to determine the time it takes for the coconut to fall and the distance the zookeeper covers in that time.
We have already calculated the time it takes for the coconut to fall (2.02 seconds) in the previous explanation.
To find the distance the zookeeper covers, we can again use the equation:
d = v * t
Plugging in the values, we have:
d = 1.5 * 2.02 = 3.03 meters
Since the zookeeper is running beneath the tree, the coconut will hit the ground 3.03 meters behind him.
3. To find the average velocity of the man from his home to the neighboring town, we need to calculate the total displacement and divide it by the total time taken. Average velocity is defined as the ratio of displacement to time.
Given that the man traveled a distance of 20 km from his home to the neighboring town, and the total time taken for the round trip was 2 hours (or 2 * 60 = 120 minutes), we can calculate the average velocity as follows:
Average velocity = Total displacement / Total time taken
Average velocity = 20 km / 2 hours = 10 km/h
Therefore, the man's average velocity from his home to the neighboring town is 10 km/h.
4. To find the resultant force, we need to take into account the direction and magnitude of the two given forces.
We have two forces: 100 N at an angle of 170 degrees and 100 N at an angle of 50 degrees.
To find the resultant force, we can use vector addition. We break each force into its respective x and y components and then add them separately.
First, let's find the x and y components of each force:
Force 1 (100 N at 170 degrees):
Fx1 = 100 N * cos(170 degrees)
Fy1 = 100 N * sin(170 degrees)
Force 2 (100 N at 50 degrees):
Fx2 = 100 N * cos(50 degrees)
Fy2 = 100 N * sin(50 degrees)
Next, add the x and y components separately:
Resultant force in the x-direction: Fx = Fx1 + Fx2
Resultant force in the y-direction: Fy = Fy1 + Fy2
Finally, we can find the magnitude and direction of the resultant force using the Pythagorean theorem and trigonometry:
Magnitude of the resultant force: Fr = sqrt(Fx^2 + Fy^2)
Direction of the resultant force: θ = atan(Fy / Fx)
Plugging in the values and performing the calculations will give you the resultant force and its direction.
5. To find the frictional force retarding the motion of the box, we can use Newton's second law of motion, 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
The net force acting on the box is the horizontal component of the applied force minus the frictional force:
Net force = Applied force - Frictional force
The horizontal component of the applied force can be found using trigonometry:
Horizontal component of the applied force = Applied force * cos(angle)
By substituting the values given in the problem, we have:
Net force = (40 N * cos(37 degrees)) - Frictional force = 5 kg * 3 m/s^2
Now, we can solve for the frictional force:
Frictional force = (40 N * cos(37 degrees)) - 5 kg * 3 m/s^2
Performing the calculations will give you the exact value of the frictional force.