A force of 5 kg - wt is acting on a body of mass 5 kg kept on a smooth horizontal surface at rest- If it acts for 10 seconds at an angle of 60- with the horizontal- find the distance travelled by the body- Take g -10 m - s 2 A- 100 mB- 150 mC- 200 mD- 250 m

The correct option is D 250 m1 kg wt = g N nbsp; Hence, force acting on body F = 5× 10 = 50 N. The component of the force along the surface = 50cos60^∘ = 25 ...

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Standard XII

Physics

Equations of Motion

A force of 5 ...

Question

A force of $5 k g - w t$ is acting on a body of mass $5 k g$ kept on a smooth horizontal surface at rest. If it acts for $10$ seconds at an angle of $60^{\circ}$ with the horizontal, find the distance travelled by the body. (Take $g = 10 m / s^{2})$

100 m

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150 m

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200 m

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250 m

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Solution

The correct option is D $250 m$
$1 k g - w t = g N$
Hence, force acting on body $F = 5 \times 10 = 50 N .$
The component of the force along the surface $= 50 cos 60^{\circ} = 25 N$

Hence acceleration $a = \frac{25}{5} = 5 m / s^{2}$
Using second equation of motion, $s = u t + \frac{1}{2} a t^{2} = 0 + \frac{1}{2} \times 5 \times 10^{2} = 250 m$

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A force of 5 kg−wt is acting on a body of mass 5 kg kept on a smooth horizontal surface at rest. If it acts for 10 seconds at an angle of 60∘ with the horizontal, find the distance travelled by the body. (Take g=10 m/s2)

The correct option is D 250 m1 kg−wt=g N Hence, force acting on body F=5×10=50 N. The component of the force along the surface =50cos60∘=25 N Hence acceleration a=255=5 m/s2 Using second equation of motion, s=ut+12at2=0+12×5×102=250 m

A force of $5 k g - w t$ is acting on a body of mass $5 k g$ kept on a smooth horizontal surface at rest. If it acts for $10$ seconds at an angle of $60^{\circ}$ with the horizontal, find the distance travelled by the body. (Take $g = 10 m / s^{2})$