Solution:
Solution
Given : $$m = 10$$ kg
Frictional force does negative work i.e $$W_f = -375$$ J
Work done by worker $$W = 500$$ J
Using work-energy theorem : $$W_{all force} = \Delta K.E$$
$$\therefore$$ $$W + W_f = \dfrac{1}{2}mv^2 - \dfrac{1}{2}mu^2$$
OR $$500 - 375 = \dfrac{1}{2}\times 10v^2 - 0$$ $$(\because u = 0 m/s)$$
$$\implies v = \sqrt{25} = 5$$ m/s