How much net work is required to accelerate a 1200 kg car from 10\(ms^{-1}\) to 15\(ms^{-1}\)
A.
1.95×\(10^5 j\)
B.
1.35×\(10^4 j\)
C.
7.5×\(10^4 j\)
D.
6.0×\(10^4 j\)
Correct answer is C
m=1200kg, \(V_1\)= \(10ms^{-1}\) \(V_2\) = \(15ms^{-1}\), w= ?
work=►K.E = \( K.E_2\) = \(K.E_2\) - \(K.E_1\)
⇒work= \(\frac{1}{2}{mv^2_2}-\frac{1}{2}{mv^2_1}\)
⇒work= \(\frac{1}{2}m({v^2_2}-{v^2_1}\))
⇒work= \(\frac{1}{2}× 1200× (15^2-10^2)\)
⇒work = 600 × (225 -100)
⇒work= 600 × 125
⇒work= 7.5×\(10^4 j\)
