Question

A small object with mass m=2 kg is attached to the free end of an ideal spring with k=10 N-m. The other end of the spring is connected to a fixed frictionless pivot located at the origin O as shown in the Figure. The relaxed Length of the spring is 1 m. An unspecified force F carries The object at rest from Point A to Point B. At point B, the object has the speed of 5 ms^-1 . the work done by the force F is

Answer 1

Answer:

The work done by unspecified force is -80 J

Explanation:

As we know that the object was at rest initially while its final speed at point B is v = 5 m/s

now we know that

$W_{ex} + W_{sp} = \Delta K$

so here we will have

$W_{sp} = \frac{1}{2}k(x_1^2 - x_2^2)$

$W_{sp} = \frac{1}{2}10((6 - 1)^2 - (3 - 1)^2)$

$W_{sp} = 5(25 - 4)$

$W_{sp} = 105 J$

now we have

$W_{ex} + 105 = \frac{1}{2}2(5^2 - 0)$

$W_{ex} + 105 = 25$

$W_{ex} = 25 - 105$

$W_{ex} = -80 J$

#Learn

Topic : Work energy Theorem

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