Chapter 10: Q. 70 (page 231)

In a physics lab experiment, a compressed spring launches a $20 g$ metal ball at a $30 °$ angle. Compressing the spring $20 c m$ causes the ball to hit the floor $1.5 m$ below the point at which it leaves the spring after traveling $5.0 m$ horizontally. What is the spring constant?

Short Answer

Expert verified

The spring constant is $19.7 N / m .$

Step by step solution

Given information

Compression of the spring $= 20 c m$

The weight of the metal ball $= 20 g$

The launching angle $= 30 °$

The horizontal distance traveled $= 5.0 m$ $= 5.0 m$

The vertical distance traveled $= 1.5 m$

Explanation

When the ball hits the floor the vertical (y) and horizontal distance (x) are 1.5 and 5 respectively. $1.5 m and5 m$ respectively.

By using the equation of motion

y = v_{y} \times t - \frac{1}{2} g t^{2} \dots \dots (1) where v_{y} isthevelocityinverticaldirection \tan (θ) = \frac{v_{y}}{v_{x}} v_{y} = v_{x} \tan (30 °) \dots \dots (1) S u b s t i t u t e e q u a t i o n (2) i n (1) 1.5 m = v_{x} \sin 30 ° \times t - \frac{1}{2} \times (9.80 m / s^{2}) \times t^{2} \to (1) H o r i z o n t a l d i s \tan c e : x = v_{x} \times t w h e r e v_{x} i s t h e v e l o c i t y i n h o r i z o n t a l d i s \tan c e . t = \frac{5.0 m}{v_{x}} \dots \dots (3) S u b s t i t u t e e q u a t i o n (3) i n (1) y = v_{x} \tan (30 °) \times \frac{5.0 m}{v_{x}} - \frac{1}{2} g t^{2} y = \tan (30 °) \times (5 m) - \frac{1}{2} g t^{2} 1.5 m = \tan (30 °) \times (5 m) - \frac{1}{2} \times 9.80 m / s^{2} \times {(\frac{5.0 m}{v_{x} \cos 30 °})}^{2} 1.5 m = \tan (30 °) \times 5 m - \frac{25 m^{2} \times 9.8 m / s^{2}}{2 \times {(v_{x} \cos 30 °)}^{2}} 1.5 m = \frac{1}{\sqrt{3}} \times 5 m - \frac{245 m^{3} / s^{2}}{2 \times {(v_{x})}^{2} \times {(\frac{\sqrt{3}}{2})}^{2}} 1.5 m = 2.885 m - \frac{245 m^{3} / s^{2}}{{(v_{x})}^{2} \times 1.5} 1.5 m = 2.855 m - \frac{163.33 m^{3} / s^{2}}{{(v_{2})}^{2}} \frac{163.33 m^{3} / s^{2}}{{(v_{x})}^{2}} = (2.855 m) - (1.5 m) {(v_{x})}^{2} = \frac{163.33 m^{3} / s^{2}}{1.5 m + 2.855 m} {(v_{x})}^{2} = 37.5 m^{2} / s^{2} v_{x} = 6.12 m / s

By using energy conservation:

$\frac{1}{2} k x^{2} = \frac{1}{2} m v^{2} x = 20 c m k (0.2 m) = (0.02 k g) {(6.12 m / s)}^{2} k = \frac{(0.02 k g) {(6.12 m / s)}^{2}}{0.2 m} k = 19.7 N / m$

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In a physics lab experiment, a compressed spring launches a $20 g$ metal ball at a $30 °$ angle. Compressing the spring $20 c m$ causes the ball to hit the floor $1.5 m$ below the point at which it leaves the spring after traveling $5.0 m$ horizontally. What is the spring constant?

Short Answer

Step by step solution

Given information

Explanation

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In a physics lab experiment, a compressed spring launches a 20gmetal ball at a 30°angle. Compressing the spring 20cmcauses the ball to hit the floor 1.5mbelow the point at which it leaves the spring after traveling 5.0mhorizontally. What is the spring constant?

Short Answer

Step by step solution

Given information

Explanation

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Geometrical and Physical Optics

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Scientific Method Physics

Study anywhere. Anytime. Across all devices.

In a physics lab experiment, a compressed spring launches a $20 g$ metal ball at a $30 °$ angle. Compressing the spring $20 c m$ causes the ball to hit the floor $1.5 m$ below the point at which it leaves the spring after traveling $5.0 m$ horizontally. What is the spring constant?