JEE MAIN - Physics (2016 - 10th April Morning Slot - No. 4)
Velocity-time graph for a body of mass 10 kg is shown in figure. Work-done on
the body in first two seconds of the motion is :
_10th_April_Morning_Slot_en_4_1.png)
_10th_April_Morning_Slot_en_4_1.png)
12000 J
$$-$$ 12000 J
$$-$$ 4500 J
$$-$$ 9300 J
Explanation
_10th_April_Morning_Slot_en_4_2.png)
Here u = 50 m/s , what t = 0
$$\alpha $$ = $${{\Delta v} \over {\Delta t}}$$ = $${{50 - 0} \over {0 - 10}}$$ = $$-$$5 m/s2
Speed of the body at t = 2 s
v = u + at
= 50 + ($$-$$ 5) $$ \times $$ 2
= 40 m/s
From work energy theorem,
$$\Delta $$w = $${1 \over 2}m{v^2} - {1 \over 2}m{u^2}$$
= $${1 \over 2}$$ m(v2 $$-$$u2)
= $${1 \over 2}$$ $$ \times $$ 10 $$ \times $$ (402 $$-$$ 502)
= 5 $$ \times $$ ($$-$$10)(90)
= $$-$$ 4500 J
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