A light spring having a force constat of 125 N/m is used to pull a 9.50kg sled on a horizontal frixtionless ice rink. If the sled has an acceleration of 2.00 m/s^2, by how much does the spring stretch if it pulls on the sled (a) horizontally, (b) at 30.0 degrees above the horizontal
f=ma
kx=ma
x= ma/k horizontally
If at 30 deg above.
x= ma/k (1/cos30)
a) Well, it looks like the spring's got some serious pulling power! To figure out how much it stretches when pulling the sled horizontally, we can use Hooke's law. The formula is F = kx, where F is the force applied, k is the spring constant, and x is the displacement. Rearranging the formula, we have x = F/k. Plugging in the values, we get x = 125 N / (125 N/m) = 1 meter. So, the spring will stretch by 1 meter when pulling the sled horizontally.
b) Now, let's kick things up a notch and pull the sled at an angle of 30 degrees above the horizontal. Don't worry, the spring won't chicken out! To find out how much it stretches this time, we need to calculate the horizontal component of the force applied by the spring. This can be done by multiplying the force applied by the cosine of the angle. So, the horizontal component of the force is 125 N * cos(30 degrees) = 108.25 N. Now we can use Hooke's law while considering only the horizontal force. Plugging in the new force, we get x = 108.25 N / (125 N/m) = 0.866 meters. So, the spring will stretch by approximately 0.866 meters when pulling the sled at a 30-degree angle above the horizontal. That's quite a stretch, huh?
To determine the amount by which the spring stretches, we need to use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position.
(a) When the spring pulls the sled horizontally, the force it exerts on the sled is equal to the force of friction acting on the sled. Since the ice rink is frictionless, the only force acting on the sled is the force from the spring.
We can use the equation F = k * x, where F is the force, k is the force constant of the spring, and x is the displacement of the spring from its equilibrium position.
Given that the force constant of the spring (k) is 125 N/m, and the mass of the sled (m) is 9.50 kg, and the acceleration of the sled (a) is 2.00 m/s², we can calculate the displacement (x).
F = k * x
ma = k * x
9.50 kg * 2.00 m/s² = 125 N/m * x
19.0 N = 125 N/m * x
x = 19.0 N / 125 N/m
x ≈ 0.152 m
Therefore, the spring stretches approximately 0.152 meters (or 15.2 cm) horizontally when it pulls on the sled.
(b) When the spring pulls the sled at an angle of 30.0 degrees above the horizontal, we need to find the horizontal component of the force exerted by the spring.
The horizontal component of the force can be calculated using the equation F_horizontal = F * cos(theta), where F is the total force exerted by the spring and theta is the angle above the horizontal.
Given that F is the same as before (19.0 N), theta is 30.0 degrees, we can calculate the horizontal component of the force.
F_horizontal = 19.0 N * cos(30.0 degrees)
F_horizontal = 19.0 N * 0.866
F_horizontal ≈ 16.43 N
Now, we can use the equation F_horizontal = k * x to find the displacement (x).
16.43 N = 125 N/m * x
x = 16.43 N / 125 N/m
x ≈ 0.131 m
Therefore, the spring stretches approximately 0.131 meters (or 13.1 cm) when it pulls on the sled at a 30.0 degrees angle above the horizontal.
To find the amount by which the spring stretches in each case, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to its displacement. The formula for Hooke's Law is given as:
F = k * x
Where:
- F is the force exerted by the spring
- k is the force constant of the spring
- x is the displacement or stretch of the spring
(a) When the spring pulls the sled horizontally, the force exerted by the spring is equal to the weight of the sled. The weight of the sled can be calculated using the formula:
Weight = mass * acceleration due to gravity
Given that the mass of the sled (m) is 9.50 kg and the acceleration due to gravity (g) is approximately 9.8 m/s^2, we can calculate the force exerted by the spring as follows:
Force (F) = Weight = m * g
Substituting the given values:
F = 9.50 kg * 9.8 m/s^2
Now, we can use Hooke's Law to find the displacement of the spring (x):
F = k * x
x = F / k
Substituting the force constant (k) as 125 N/m and the force (F) calculated above, we can find the stretch of the spring (x) when pulled horizontally.
(b) When the spring pulls the sled at 30.0 degrees above the horizontal, the force exerted by the spring can be decomposed into two components: one along the horizontal direction and one along the vertical direction. The horizontal component of the force will contribute to the stretch of the spring, while the vertical component will not affect the stretch.
The horizontal component of the force can be calculated using trigonometry. Given that the force exerted by the spring is F, we can find the horizontal component (F_h) as follows:
F_h = F * cos(theta)
Where theta is the angle above the horizontal, which is 30.0 degrees in this case.
Now, we can use Hooke's Law again to find the displacement of the spring (x) using the horizontal component of the force (F_h):
F_h = k * x
x = F_h / k
Substituting the force constant (k) as 125 N/m and the horizontal component of the force (F_h) calculated above, we can find the stretch of the spring (x) when pulled at 30.0 degrees above the horizontal.