Open Question
A 950-kg car strikes a huge spring at a speed of 25 m/s (Fig. 14–43), compressing the spring 4.0 m.
(b) How long is the car in contact with the spring before it bounces off in the opposite direction?
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17. Periodic Motion
Spring Force (Hooke's Law)
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A20.0 cmmassless spring with spring constant340 N/mis suspended from the ceiling. A student carefully hangs a450 gmass from the free end. How long is the spring now?
A
18.7 cm
B
20.0 cm
C
24.5 cm
D
26.7 cm
E
21.3 cm
Verified step by step guidance1
Identify the given values: the spring constant \( k = 340 \text{ N/m} \), the mass \( m = 450 \text{ g} = 0.450 \text{ kg} \), and the original length of the spring \( L_0 = 20.0 \text{ cm} = 0.200 \text{ m} \).
Use Hooke's Law to find the extension of the spring. Hooke's Law is given by \( F = kx \), where \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the extension of the spring.
Calculate the force exerted by the mass due to gravity using \( F = mg \), where \( g = 9.8 \text{ m/s}^2 \) is the acceleration due to gravity. Substitute the values to find \( F \).
Rearrange Hooke's Law to solve for the extension \( x \): \( x = \frac{F}{k} \). Substitute the calculated force and the spring constant to find \( x \).
Add the extension \( x \) to the original length of the spring \( L_0 \) to find the new length of the spring: \( L = L_0 + x \). Convert the result back to centimeters if necessary.
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