a 30.0 kg object has an initial velocity of 7.00 m/s and a final velocity of 10.0 m/s. the work done on the…

Answer:

$645$

Explanation:

Step1: Recall work - energy theorem

Work done $W=\Delta K = K_f - K_i$

Step2: Calculate initial kinetic energy

$K_i=\frac{1}{2}mv_i^2$, where $m = 30.0\ kg$ and $v_i=7.00\ m/s$. So $K_i=\frac{1}{2}\times30.0\times(7.00)^2=\frac{1}{2}\times30.0\times49.0 = 735\ J$

Step3: Calculate final kinetic energy

$K_f=\frac{1}{2}mv_f^2$, where $m = 30.0\ kg$ and $v_f = 10.0\ m/s$. So $K_f=\frac{1}{2}\times30.0\times(10.0)^2=\frac{1}{2}\times30.0\times100=1500\ J$

Step4: Calculate work done

$W=K_f - K_i=1500 - 735=765\ J$