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2 years ago

A 0.54 kg particle has a speed of 5.0 m/s at point A and kinetic energy of 7.5 J at point B. What is

Physics

1 answer:

Wewaii [24]2 years ago

4 0

Answers

(a) 6.75 Joules.

(b) 5.27 m/s

Explanation

Kinetic energy is the energy possessed by a body in motion.

(a) its kinetic energy at A?

K.E = 1/2 mv²

= 1/2 × 0.54 × 5²

= 6.75 Joules.

(b) its speed at point B?

K.E = 1/2 mv²

7.5 = 1/2 × 0.54 × V²

V² = 7.5 ÷ 0.27

= 27.77778

V = √27.77778

= 5.27 m/s

Work done = 7.5 - 6.75

= 0.75 Joules

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$\mathbf x(t)=\displaystyle\int_0^t\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf j\,\mathrm du$

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$\mathbf x(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)t\,\mathbf i+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)t^2\,\mathbf j$

Get the coordinates at t = 8.00 s by evaluating $\mathbf x(t)$ at this time:

$\mathbf x(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)(8.00\,\mathrm s)\,\mathbf i+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)(8.00\,\mathrm s)^2\,\mathbf j$

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so the particle is located at (x, y) = (64.0, 64.0).

Get the speed at t = 8.00 s by evaluating $\mathbf v(t)$ at the same time:

$\mathbf v(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)(8.00\,\mathrm s)\,\mathbf j$

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