Question 1:
Suppose a child of weight w climbs onto the sled. If the tension force is measured to be 58.5 N, find the weight of the child and the magnitude of the normal force acting on the sled.
Question 2 :
(a) Suppose a hockey puck slides down a frictionless ramp with an acceleration of 4.50 m/s2. What angle does the ramp make with respect to the horizontal?
°
(b) If the ramp has a length of 5.80 m, how long does it take the puck to reach the bottom?
s
(c) Now suppose the mass of the puck is doubled. What's the puck's new acceleration down the ramp?
m/s2
Question 1:
We are given the tension force T=58.5 N and we need to find the weight(w) of the child and the normal force(N) acting on the sled.
Using Newton's second law, we can write the equations for the forces acting along the parallel(perpendicular) direction to the inclined plane:
T - w*sin(theta) = 0
N - w*cos(theta) = 0
Here, theta is the angle between the inclined plane and the horizontal. It is given that theta = 30°.
Using the first equation, we can find the weight of the child:
w*sin(30) = 58.5
w = 58.5/sin(30)
w = 58.5/0.5
w = 117 N
Therefore, the weight of the child is 117 N.
Now, using the second equation to find the magnitude of the normal force acting on the sled:
N = w*cos(30)
N = 117*cos(30)
N = 117*0.866
N ≈ 101.3 N
Therefore, the normal force acting on the sled is approximately 101.3 N.
Question 2:
(a) To find the angle the ramp makes with respect to the horizontal, we use the formula for acceleration along an inclined plane without friction:
a = g*sin(theta)
Where a is the acceleration, g is the acceleration due to gravity (9.81 m/s^2), and theta is the angle. We are given a=4.50 m/s^2.
theta = arcsin(a/g)
theta = arcsin(4.50/9.81)
theta ≈ 27.5°
The angle the ramp makes with respect to the horizontal is approximately 27.5°.
(b) To find the time it takes the puck to reach the bottom, we can use the formula:
s = 0.5*a*t^2
Where s is the length of the ramp, a is the acceleration, and t is the time. We are given s = 5.80 m and a = 4.50 m/s^2.
5.80 = 0.5 * 4.50 * t^2
t^2 = 5.80/(0.5 * 4.50)
t^2 ≈ 2.58
t ≈ 1.61 s
It takes the puck approximately 1.61 seconds to reach the bottom of the ramp.
(c) If the mass of the puck is doubled, the acceleration down the ramp will remain the same. This is because the gravitational force acting on the puck is directly proportional to its mass, so doubling the mass will double the gravitational force. However, the normal force acting on the puck will also double. Since these two forces are balanced, the acceleration remains unchanged. Thus, the new acceleration down the ramp is the same as before:
New acceleration = 4.50 m/s^2
Question 1:
To find the weight of the child, we can use the formula:
Weight = Tension force
Weight = 58.5 N
So, the weight of the child is 58.5 N.
To find the magnitude of the normal force acting on the sled, we can use the equation:
Net force = Weight + Normal force
Since the sled is not accelerating vertically, the net force is equal to zero. Therefore:
Weight + Normal force = 0
Substituting the weight we just found:
58.5 N + Normal force = 0
Solving for the magnitude of the normal force:
Normal force = -58.5 N (since it is acting in the opposite direction)
So, the magnitude of the normal force acting on the sled is 58.5 N in the opposite direction.
Question 2:
(a) To find the angle the ramp makes with the horizontal, we can use the equation:
Acceleration = g * sin(angle)
Where "g" is the acceleration due to gravity (approximated as 9.8 m/s^2).
Given:
Acceleration = 4.50 m/s^2
g = 9.8 m/s^2
Rearranging the equation:
sin(angle) = Acceleration / g
sin(angle) = 4.50 m/s^2 / 9.8 m/s^2
Using a scientific calculator, find the inverse sine (sin^-1) of the result to get the angle in radians. Then, convert the angle from radians to degrees.
The angle the ramp makes with the horizontal is approximately °.
(b) To find the time it takes for the puck to reach the bottom of the ramp, we can use the equation:
Distance = (1/2) * Acceleration * Time^2
Given:
Acceleration = 4.50 m/s^2
Distance = 5.80 m
Rearranging the equation:
Time^2 = (2 * Distance) / Acceleration
Time = √((2 * 5.80 m) / 4.50 m/s^2)
Calculate the square root of the result to find the time.
The puck takes approximately seconds to reach the bottom of the ramp.
(c) If the mass of the puck is doubled, the force acting on it will also double due to Newton's second law (F = m * a). Therefore, the puck's new acceleration down the ramp will remain the same: 4.50 m/s^2.
Question 1:
To find the weight of the child and the magnitude of the normal force acting on the sled, we can use Newton's second law of motion and the concept of equilibrium.
Newton's second law states that the sum of all forces acting on an object is equal to the mass of the object multiplied by its acceleration. In this case, since the sled and the child are in equilibrium (not accelerating), the sum of the forces must be zero.
Let's break down the forces acting on the sled and the child. We have the weight (mg) acting vertically downward and the tension force (T) acting vertically upward. The normal force (N) acts perpendicularly to the surface the sled is on.
Since the sled is not moving vertically, the sum of the vertical forces must be zero. Therefore, we can say:
T - mg - N = 0
Now, we know the tension force (T) is 58.5 N, but we need to find the weight of the child (mg) and the magnitude of the normal force (N).
To find the weight of the child, we can rearrange the equation above as:
mg = T - N
To find the magnitude of the normal force, we can rearrange the equation again:
N = T - mg
So, to find the weight of the child, we can substitute the value of T into the equation, and to find the magnitude of the normal force, we can substitute both T and the weight (mg) into the equation.
Question 2:
(a) To find the angle the ramp makes with respect to the horizontal, we can use the formula for the acceleration of an object sliding down a ramp. The formula is:
acceleration = g * sin(theta)
Where g is the acceleration due to gravity (9.8 m/s^2) and theta is the angle the ramp makes with the horizontal. We are given the acceleration (4.50 m/s^2). Rearranging the formula, we can solve for theta:
theta = arcsin(acceleration / g)
(b) To find how long it takes the puck to reach the bottom of the ramp, we can use the kinematic equation for motion along an inclined plane. The equation is:
distance = (initial velocity * time) + (0.5 * acceleration * time^2)
Since the puck starts from rest (initial velocity = 0) and the distance is given (5.80 m), we can rearrange the equation to solve for time:
time = sqrt((2 * distance) / acceleration)
(c) To find the new acceleration of the puck down the ramp when its mass is doubled, we can use Newton's second law of motion. The formula is:
force = mass * acceleration
The force acting on the puck is its weight (mg), and since the mass is doubled, the new force will be twice the original force. Rearranging the formula, we can solve for the new acceleration:
new acceleration = force / (2 * mass)