Answered

Westonci.ca makes finding answers easy, with a community of experts ready to provide you with the information you seek. Our platform provides a seamless experience for finding reliable answers from a knowledgeable network of professionals. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.

The picture below shows a block of mass 5.22 kg held against a spring of spring constant k compressed by a distance d = 28.3 cm on a frictionless ramp of length L. The ramp has an angle of 20∘. A distance x = 1.16 m separates the ramp from a landing area that is at the same height as the top of the ramp.


A) If the mass leaves the end of the spring at a speed of 4.43 m/s , what is the spring constant of the spring? k = ____ N/m

B) If the mass reaches the top of the ramp moving at speed 2.81 m/s , what is the length of the ramp?

Measure the length of the ramp from the starting point (compressed spring) as shown in the picture. L = ____ m

C) Does the mass successfully jump the gap and land on the other side?

D) At what height h below the top of the ramp must the landing area be placed for the block to successfully clear the 1.16 m wide gap? h= ___ m

Sagot :

onyebuchinnaji

(a) The spring constant is 1,279.1 N/m.

(b) The length of the ramp is 1.75 m.

(c) The mass did not jump the gap.

(d)  The landing area must be placed 1.98 m below the top of the ramp for the block to successfully jump it.

Spring constant

The spring constant is calculated as follows;

¹/₂kx² = ¹/₂mv²

kx² = mv²

k = mv²/x²

k = (5.22 x 4.43²) / (0.283)²

k = 1,279.1 N/m

Displacement of the mass

The height traveled by the mass when it reaches top of the ramp is calculated as follows;

¹/₂Kd² = ¹/₂mv² + mgh

The height of the ramp, h = Lsinθ

¹/₂Kd² = ¹/₂mv² + mgLsinθ

0.5 x 1279.1 x 0.283² = 0.5 x 5.22 x (2.81)² + 5.22 x 9.8 x (sin20) x L

51.22 = 20.61 + 17.5L

30.61 = 17.5L

L = 30.61 / 17.5

L = 1.75 m

Height of the ramp

h = L x sin20

h = 1.75 x sin20

h = 0.6 m

Time taken for the mass to travel the height

[tex]h = vt + \frac{1}{2} gt^2\\\\0.6 = 2.81t + 4.9t^2\\\\4.9t^2 + 2.81t - 0.6=0\\\\t = 0.17 \ s[/tex]

Horizontal distance = vt = 0.17 x 2.81 = 0.48 m (0.48 m is less than 1.16 m)

Thus, the mass did not jump the gap.

Time of motion if the mass must jump 1.16 m

t = x/v

t = 1.16/2.81

t = 0.41

Height it must be placed

[tex]H = 2.81(0.41) + \frac{1}{2} (9.8) (0.41)^2\\\\H = 1.98 \ m[/tex]

Thus, the landing area must be placed 1.98 m below the top of the ramp for the block to successfully jump it.

Learn more about conservation of energy here: https://brainly.com/question/166559

Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Thank you for trusting Westonci.ca. Don't forget to revisit us for more accurate and insightful answers.