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A 1-kg block is loaded into a mechanism to send it along a set of rails as shown

ID: 1288839 • Letter: A

Question

A 1-kg block is loaded into a mechanism to send it along a set of rails as shown in the figure below. The spring mechanism has a spring constant of 3.6x103 N/m and has a frictionless surface. The block is slowly pressed against spring compressing it and then the block is held motionless with a force of 400 N. The block is then released sending it along rails that have a coefficient of static friction of 0.6 and kinetic friction of 0.5. After distance 100 cm the rails bend upward at an angle of 30° and end after an additional length of 100 cm.

a. Determine how far the spring is compressed.
b. Determine the speed of the block as it exits the mechanism and starts along the track.
c. Determine the speed of the block when it first encounters the ramp.
d. Determine the maximum height the block reaches.
e. Find the where the block hits the ground measured from the x = 0 position marked in the diagram.

A 1-kg block is loaded into a mechanism to send it along a set of rails as shown in the figure below. The spring mechanism has a spring constant of 3.6x103 N/m and has a frictionless surface. The block is slowly pressed against spring compressing it and then the block is held motionless with a force of 400 N. The block is then released sending it along rails that have a coefficient of static friction of 0.6 and kinetic friction of 0.5. After distance 100 cm the rails bend upward at an angle of 30 A hat A degree and end after an additional length of 100 cm. a. Determine how far the spring is compressed. b. Determine the speed of the block as it exits the mechanism and starts along the track. c. Determine the speed of the block when it first encounters the ramp. d. Determine the maximum height the block reaches. e. Find the where the block hits the ground measured from the x = 0 position marked in the diagram

Explanation / Answer

a>. kx = 400N

x = 400/3.6 * 1000

=11.11 cm

b>. 1/2 k x2 = 1/2 m v2

v2 =43.56

v = 6.6 m/s

c>. frictional force (f)= 0.5 * 1 * 9.8 N = 4.9 N

By Energy balance

1/2mu2 = 1/2 m v2 +f ( d + x) ; u =6.6m/s

v2 = 32.68

v =5.7m/s

d>. (mg sin@ + 0.5 mg cos@ ) * L = 1/2 m v^2 ; @ = 30

L = 1.77 m

e>. on the inclined plane

(mgsin@ - 0.5 mgcos@ ) * L + mgh= 1/2 m v^2 ; h = L sin@

v = 4.43m/s

on the plane

0.5 mg (d + x) + 1/2 kx^2 = 1/2 m v^2

x = 0.05 m