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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 :

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(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

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