5. An 0.80-kg block is held in place against the spring by a 67-N horizontal ext

Question

5. An 0.80-kg block is held in place against the spring by a 67-N horizontal external force (see the figure). The external force is removed, and the block is projected with a velocity Vi = 1.2 m/s upon separation from the spring. The block descends a ramp and has a velocity V2 = 1.9 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 over this section is 0.39. The velocity of the block is V3 = 1.4 m/s at C. The block moves on to D, where it stops. What is the spring constant of the spring? 6. In the figure, a block of mass m is moving along the horizontal frictionless surface with a speed of 5.70 m/s. If the slope is 11.00 and the coefficient of kinetic friction between the block and the incline is 0.2 00, how far does the block travel up the incline?

Explanation / Answer

5) Apply, Hook's law

F = k*x

67 = k*x ----(1)

Apply conservation of energy

initial poetntail energy in the spring = kinetic energy of the block when the block leaves the spring.

0.5*k*x^2 = 0.5*m*v^2

(k*x)*x = m*v^2

F*x = m*v^2

x = m*v^2/F

= 0.8*1.2^2/67

= 0.0172 m

now From equation 1

F = k*x

==> k = F/x

= 67/0.0172

= 3897 N/m <<<<<<<----------Answer

6) acceleration of the block, a = -g*sin(11) - mue_k*g*cos(11)

= -9.8*sin(11) - 0.2*9.8*cos(11)

= -3.794 m/s^2

u = 5.7 m/s (initial speed)

v = 0 (final speed)

Let d is the distance travelled along the ramp before coming to stop.

Apply, d = (v^2 - u^2)/(2*a)

= (0^2 - 5.7^2)/(2*(-3.794)

= 4.28 m <<<<<<<----------Answer

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