An ideal spring is mounted horizontally, with its left end fixed. The force constant of the spring is 170 N/m. A glider of mass

Answer:

The pen requires 7.2 mJ of energy to extend.

Explanation:

Provided:

Length = 1.8 cm

Spring constant = 300 N/m

Initial compression = 1.0 mm

Additional compression = 6.0 mm

Total compression = 1.0 + 6.0 = 7.0 mm

We need to determine the energy needed

This energy is equivalent to the variation in spring potential energy

Substitute the values into the formula

Therefore, a total of 7.2 mJ is needed to extend the pen.

The question lacks details. Here is the full question.

The accompanying image was captured with a camera capable of shooting between one and two frames per second. A series of photos was merged into this single image, meaning the vehicles depicted are actually the same car, documented at different intervals.

Assuming the camera produced 1.3 frames per second for this image and that the length of the car is approximately 5.3 meters, based on this information and the photo, how fast was the car moving?

Answer: v = 6.5 m/s

Explanation: The problem requires calculating the car's velocity. Velocity can be computed using:

Since the camera captured 7 images of the car and its length is noted as 5.3, the car's displacement is:

Δx = 7(5.3)

Δx = 37.1 m

The camera operates at 1.3 frames per second and recorded 7 images, thus the time driven by the car is:

1.3 frames = 1 s

7 frames = Δt

Δt = 5.4 s

v = 6.87 m/s

Case A:

A.75 kg 65 N/m 1.2 m

m = weight of the car = 0.75 kg

k = spring's stiffness = 65 N/m

h = elevation of the hill = 1.2 m

x = spring's compression = 0.25 m

Applying the principle of energy conservation from the Top of the hill to the Bottom of the hill

Energy at the Top of the hill equals Energy at the Bottom of the hill

spring energy + gravitational potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (65) (0.25)² + (0.75 x 9.8 x 1.2) = (0.5) (0.75) v²

v = 5.4 m/s

Case B:

B.60 kg 35 N/m.9 m

m = weight of the car = 0.60 kg

k = spring's stiffness = 35 N/m

h = height of the hill = 0.9 m

x = spring's compression = 0.25 m

Using energy conservation from the Top of the hill to the Bottom of the hill

Top hill energy = Bottom hill energy

spring energy + gravitational potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (35) (0.25)² + (0.60 x 9.8 x 0.9) = (0.5) (0.60) v²

v = 4.6 m/s

Case C:

C.55 kg 40 N/m 1.1 m

m = weight of the car = 0.55 kg

k = spring's stiffness = 40 N/m

h = height of the hill = 1.1 m

x = spring's compression = 0.25 m

Using conservation of energy from the Top of the hill to the Bottom of the hill

Energy at the Top of the hill = Energy at the Bottom of the hill

spring energy + gravitational potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (40) (0.25)² + (0.55 x 9.8 x 1.1) = (0.5) (0.55) v²

v = 5.1 m/s

Case D:

D.84 kg 32 N/m.95 m

m = weight of the car = 0.84 kg

k = spring's stiffness = 32 N/m

h = height of the hill = 0.95 m

x = spring's compression = 0.25 m

Using energy conservation from the Top of the hill to the Bottom of the hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + gravitational potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (32) (0.25)² + (0.84 x 9.8 x 0.95) = (0.5) (0.84) v²

v = 4.6 m/s

thus, the closest result is from case C at 5.1 m/s

Explanation:

The entire system will accelerate due to the applied force. The box will feel a force opposing friction, and once this force surpasses the friction, the box will start moving. Therefore,

Ff = μs×m1×g

m1×a = μs×m1×g

a = μs×g

The force applied is expressed as

F = (m1 + m2)×a hence

F = μs×g×(m1+m2)