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Standard XII

Physics

Relation between Work Done and PE

A mass of 1 k...

Question

A mass of 1 kg is acted upon by a single force $\to F = (4^i + 4^j) N .$
Under this force it is displaced from (0, 0 ) to (1 m, 1 m). If initially the speed of the particle was 2 m/s, its final speed should be

6 m/s

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4.5 m/s

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8 m/s

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4 m/s

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Solution

The correct option is B 4.5 m/s
Work done = change in kinetic energy.
$∴ \int_{(0, 0)}^{(1.1)} (4^i + 4^j) . (d x^i + d y^j) = \frac{1}{2} \times 1 \times (v_{f}^{2} - 2^{2})$
or $8 = \frac{1}{2} (v_{f}^{2} - 4)$
or $v_{f} = 4.5 m / s$

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A mass of 1 kg is acted upon by a single force →F=(4^i+4^j) N. Under this force it is displaced from (0, 0 ) to (1 m, 1 m). If initially the speed of the particle was 2 m/s, its final speed should be

The correct option is B 4.5 m/sWork done = change in kinetic energy. ∴∫(1.1)(0,0)(4^i+4^j).(dx ^i+dy ^j)=12×1×(v2f−22) or 8=12(v2f−4) or vf=4.5 m/s

A mass of 1 kg is acted upon by a single force $\to F = (4^i + 4^j) N .$
Under this force it is displaced from (0, 0 ) to (1 m, 1 m). If initially the speed of the particle was 2 m/s, its final speed should be

The correct option is B 4.5 m/s
Work done = change in kinetic energy.
$∴ \int_{(0, 0)}^{(1.1)} (4^i + 4^j) . (d x^i + d y^j) = \frac{1}{2} \times 1 \times (v_{f}^{2} - 2^{2})$
or $8 = \frac{1}{2} (v_{f}^{2} - 4)$
or $v_{f} = 4.5 m / s$