1
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 9th January Evening Slot

A force acts on a 2 kg object so that its position is given as a function of time as x = 3t2 + 5. What is the work done by this force in first 5 seconds ?
A
850 J
B
950 J
C
875 J
D
900 J

Explanation

Displacement,

x = 3t2 + 5

$$ \therefore $$  v = $${{dx} \over {dt}} = 6t$$

At t = 0,   velocity = 6 $$ \times $$ 0 = 0

at t = 5, velocity = 5 $$ \times $$ 6 = 30 m/s

we know from work energy theorem,

Work (W) = change in kinetic energy ($$\Delta $$K)

= $${1 \over 2}mv_F^2 - {1 \over 2}mv_i^2$$

= $${1 \over 2}$$ $$ \times $$ 2 $$ \times $$ (30)2 $$-$$ 0

= 900 J
2
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 10th January Morning Slot

A block of mass m is kept on a platform which starts from rest with constant acceleration g/2 upward, as shown in figure. Work done by normal reaction on block in time is -

A
$${{m{g^2}{t^2}} \over 8}$$
B
$${{3m{g^2}{t^2}} \over 8}$$
C
$$-$$ $${{m{g^2}{t^2}} \over 8}$$
D
0

Explanation

N $$-$$ mg = $${{mg} \over 2}$$ $$ \Rightarrow $$ N = $${{3mg} \over 2}$$

The distance travelled by the system in time t is

S = ut + $${1 \over 2}a{t^2} = 0 + {1 \over 2}\left( {{g \over 2}} \right){t^2} = {1 \over 2}{g \over 2}{t^2}$$

Now, work done

W = N.S = $$\left( {{3 \over 2}mg} \right)\left( {{1 \over 2}{g \over 2}{t^2}} \right)$$

$$ \Rightarrow $$  W = $${{3m{g^2}{t^2}} \over 8}$$
3
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 10th January Evening Slot

A particle which is experiencing a force, given by $$\overrightarrow F = 3\widehat i - 12\widehat j,$$ undergoes a displacement of $$\overrightarrow d = 4\overrightarrow i $$ particle had a kinetic energy of 3 J at the beginning of the displacement, what is its kinetic energy at the end of the displacement ?
A
9 J
B
10 J
C
12 J
D
15 J

Explanation

Work done = $$\overrightarrow F \cdot \overrightarrow d $$  

                    $$=$$ 12 J

work energy theorem

wnet $$=$$ $$\Delta $$K.E.

12 $$=$$ Kf $$-$$ 3

Kf = 15 J
4
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 11th January Morning Slot

A body of mass 1 kg falls freely from a height of 100 m, on a platform mass 3 kg which is mounted on a spring having spring constant k = 1.25 $$ \times $$ 106 N/m. The body sticks to the platform and the spring's maximum compression is found to be x. Given that g = 10 ms–2 , the value of x will be close to :
A
8 cm
B
4 cm
C
40 cm
D
80 cm

Explanation

velocity of 1 kg block just before it collides with 3kg block

= $$\sqrt {2gh} = \sqrt {2000} $$ m/s

Applying momentum conversation just before and just after collision.

1 $$ \times $$ $$\sqrt {2000} $$ = 4v $$ \Rightarrow $$ v = $${{\sqrt {2000} } \over 4}$$ m/s



initial compression of spring

1.25 $$ \times $$ 106 x0 = 30 $$ \Rightarrow $$ x0 $$ \approx $$ 0

applying work energy theorem,

Wg + Wsp = $$\Delta $$KE

$$ \Rightarrow $$   40 $$ \times $$ x + $${1 \over 2}$$ $$ \times $$ 1.25 $$ \times $$ 106 (02 $$-$$ x2)

= 0 $$-$$ $${1 \over 2}$$ $$ \times $$ 4 $$ \times $$ v2

solving x $$ \approx $$ 4 cm

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