Question

A company designs a spring system for loading ice blocks onto a truck. The ice block is placed in a holder h in front of the spring and an electric motor compresses the spring by pushing h to the left. When the spring is released the ice block is accelerated towards a ramp abc. When the spring is fully decompressed, the ice block loses contact with the spring at a. The mass of the ice block is 55 kg.

Answer 1

Answer:

Speed of block at point A is 4.93 m/s and elastic energy stored in the spring is 670 J

Explanation:

As we know that the speed of the block of ice when it reaches to the top of the truck at height 1.2 m is 0.90 m/s

so we will have

$K_c = \frac{1}{2}mv_c^2$

$K_c = \frac{1}{2}(55)(0.90)^2$

$K_c = 22.27 m/s$

now we know by mechanical energy conservation law

$K_a - K_c = mgh$

$\frac{1}{2}mv_a^2 - \frac{1}{2}mv_c^2 = mg(1.2)$

$v_a^2 - (0.90)^2 = 2(9.81)(1.2)$

$v_a = 4.93 m/s$

Now we know by energy conservation

Elastic energy stored in spring = kinetic energy of block at A

so we have

$E = \frac{1}{2}mv_a^2$

$E = \frac{1}{2}(55)(4.93)^2$

$E = 670 J$

#Learn

Topic : Elastic energy of spring

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