Question

what power is expended by a weightlifter when lifting a 66 - kg barbell a vertical distance of 1.2 m in a time interval of 1.3 s? express your answer to two significant figures and include the appropriate units. p = value units

Explanation:

Step1: Calculate the work done

Work \(W = Fd\), and the force \(F = mg\) (where \(m = 66\ kg\), \(g= 9.8\ m/s^{2}\), \(d = 1.2\ m\)). So \(F=mg=66\times9.8\ N\), and \(W = Fd=(66\times9.8)\times1.2\ J\).
\[W=66\times9.8\times1.2=66\times11.76 = 776.16\ J\]

Step2: Calculate the power

Power \(P=\frac{W}{t}\), with \(W = 776.16\ J\) and \(t = 1.3\ s\). So \(P=\frac{776.16}{1.3}\ W\).
\[P=\frac{776.16}{1.3}\approx597.05\ W\]
Rounding to two - significant figures, \(P = 6.0\times10^{2}\ W\)

Answer:

\(6.0\times 10^{2}\ W\)