Question

As shown in the figure, a 2.00-kg block is held in place against the spring by a 50-N horizontal external force. The external force is removed, and the block is projected with a velocity v1 = 1.2 m/s at as it separates from the spring. The block descends a ramp and has a velocity v2 = 2.0 m/s at the bottom. The track is frictionless between points A and B. The block enters a rough section at B, extending to E. The coefficient of kinetic friction between the block and the rough surface is 0.30. The velocity of the block is v3 = 1.4 m/s at C. The block moves on to D, where it stops.
A) What is the total energy of the spring block system before the block descends down the ramp.

B) How high is the ramp?

C) How far did the block move from B to C?

D) How far did the block move from C to D?

E) What is the work done by friction on the block from B to D?

F) What is the Mechanical energy of the spring-block system at E?

G) How much was the spring compressed by the external force

H) What is the spring constant?

Answer 1

Answer:

Explanation:

A ) Total energy of spring block system at height = kinetic energy at the bottom

= 1/2 m V²

= .5 x 2 x 2²

= 4 J

B )

height of ramp be h

mgh = 1/2 m ( v₂² - v₁² )

2 x 9.8 x h = 2² - 1.2²

19.6 h = 4 - 1.44

h = .1306 m

13.06 cm

C ) If distance between B and C be x₁

work done by friction = reduction in kinetic energy

= μ mg x₁ = 1/2 m ( 2² - 1.4² )

.3 x 9.8 x₁ = 1.02

x₁ = 34.7 cm

D )

If distance between C and D be x₂

work done by friction = reduction in kinetic energy

= μ mg x₂ = 1/2 m ( 1.4² - 0 )

.3 x 9.8 x₂ = .98  

x₂ = 33.33 cm

E )

Total work done by friction

= 1/2 m v² - 0 , v = 2 m /s

= .5 x 2 x 2²

= 4 J .

F )

Mechanical energy of the system at E = 0

G ) 1/2 k x² = 1/2 m v²

kx = 50 N , where x is compression and k is spring constant

k²x² / 2k = 1/2 m v²

50² / k = 2 x 1.2²

k = 868.05 N / m

H ) kx = 50

x = 50 / k

= 50 / 868.05

= 5.76 cm .