ICSE Class 10 Physics Question 13 of 23

Work, Energy and Power — Question 18

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Question

Question 18

A truck weighing 1000 kg changes its speed from 36 km h-1 to 72 km h-1 in 2 minutes.

Calculate :

(i) the work done by the engine and

(ii) its power

(g =10 m s-2)

Answer

Given,

Mass = 1000 kg

Initial velocity (u) =36km/h=36×1000m3600sInitial velocity (u) =10ms1\text {Initial velocity (u) } = 36 km / h \\[0.5em] = \dfrac {36 \times 1000m}{3600s} \\[0.5em] \Rightarrow \text {Initial velocity (u) } = 10ms^{-1} \\[0.5em]

Final velocity (v) =72km/h=72×1000m3600sFinal velocity (v) =20ms1\text {Final velocity (v) } = 72 km / h \\[0.5em] = \dfrac {72 \times 1000m}{3600s} \\[0.5em] \Rightarrow \text {Final velocity (v) } = 20ms^{-1} \\[0.5em]

From, equation of motion we get

s=(v2u2)/2as = ( v^2 - u^2) / 2a

(i) Work done = force x displacement

W=F.s=ma.s=1000×(202102)2W=1000×(400100)2W=500×300W=150000J=1.5×105JW = F.s = ma.s = \dfrac{1000 \times (20^2 -10^2)}{2} \\[0.5em] W = \dfrac{1000 \times (400 -100)}{2} \\[0.5em] W = 500 \times 300 \\[0.5em] \Rightarrow W = 150000J = 1.5 × 10^5J \\[0.5em]

(ii) Power = work done / time taken

Power=1.5×105120Power=1.25×103WPower = \dfrac{1.5 \times 10^5}{120} \\[0.5em] \Rightarrow Power = 1.25 × 10^3 W