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A 15 kg mass is moving at 7.50 m/s on a horizontal, frictionless surface. What is the total work that must be done on the mass to increase its speed to 11.5 m/s?

Answer :

Explanation:

From crook energy theorem

Work done = change in KE

[tex]\\\begin{aligned}&=\frac{1}{2} m u^{2}-\frac{1}{2} m v_{i}^{2} \\&=\frac{1}{2} m\left(v_{f}^{2}-v_{i}^{2}\right) \\&=0.5 \times 15.5\left(1105-7.5^{2}\right) \\w &=[589 \mathrm{J}]\end{aligned}$[/tex]

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