Question

A truck weighing 1000 kgf changes its speed from
36 km h-1 to 72 km h-1 in 2 minutes. Calculate :
(i) the work done by the engine
(ii) its power​

Answer 1

Given :

Weight of truck = 1000 kgf

Initial velocity = 36km/hr = 10m/s

Final velocity = 72km/hr = 20m/s

Time interval = 2 min = 120s

To Find :

• Work done
• Power

Solution :

❖ As per work - kinetic energy theorem,

• W = ∆k

» W denotes work done

» ∆k denotes change in kinetic energy

➙ W = ∆k = $\sf{k_f-k_i}$

➙ W = 1/2 m (v² - u²)

• Mass = 1000/10 = 100kg

➙ W = 1/2 (100) [20² - 10²]

➙ W = 50 (400 - 100)

➙ W = 50 × 300

➙ W = 15000 J = 15 kJ

We know that power is defined as the rate of work done per unit time$.$

➠ Power = Work / Time

➠ P = 15000/120

➠ P = 1500/12

➠ P = 125 W

Answer 2

Answer:

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