5.10 A body of mass 0.40 kg moving initially with a constant speed of 10 m s-1 to the north is subject to a constant force of 8.0 N directed towards the south for 30 s. Take the instant the force is applied to be t = 0, the position of the body at that time to be x = 0, and predict its position at t = –5 s, 25 s, 100 s.

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Vishal Baghel | Contributor-Level 10

4 months ago

Given, the mass of the body = 0.4 kg, Initial speed = 10m/s, Initial force, Retarding force, F = - 8 N

(a) At t = -5s, Due to constant speed, the acceleration of the body = 0

From the relation, s = ut + (1/2)at², we get

s = 10 $\times (- 5)$ + 0 $\times$ ${(\frac{1}{2}) (- 5)}^{2}$ = -50 m

At t = 25s, The acceleration of the body due to retarding force, from the relation F = ma, a = -8/0.4 m/s2 = - 20 m/s2

From the relation, s = ut + at², , we get

s = 10 $\times (25)$ + ${(\frac{1}{2}) (- 20) (- 25)}^{2}$ = -6000m

At t = 100s

For the first 30 s, while the body moves towards North, the opposite force of 8 N acts on it and from the relation s = ut + (1/2)at², , we get

S₃₀ = 10 $\times$ 30 +&nb

...more

Given, the mass of the body = 0.4 kg, Initial speed = 10m/s, Initial force, Retarding force, F = - 8 N

(a) At t = -5s, Due to constant speed, the acceleration of the body = 0

From the relation, s = ut + (1/2)at², we get

s = 10 $\times (- 5)$ + 0 $\times$ ${(\frac{1}{2}) (- 5)}^{2}$ = -50 m

At t = 25s, The acceleration of the body due to retarding force, from the relation F = ma, a = -8/0.4 m/s2 = - 20 m/s2

From the relation, s = ut + at², , we get

s = 10 $\times (25)$ + ${(\frac{1}{2}) (- 20) (- 25)}^{2}$ = -6000m

At t = 100s

For the first 30 s, while the body moves towards North, the opposite force of 8 N acts on it and from the relation s = ut + (1/2)at², , we get

S₃₀ = 10 $\times$ 30 + $(\frac{1}{2}) (- 20) 30^{2}$ = -8700m

The speed after 30 sec, v = u+at

v = 10 – (20 $\times 30)$ = -590 m/s

The distance covered in next 70 s,

S₇₀ = -590 $\times 70 + (\frac{1}{2}) (* (0) () (70)) \times (70)$ = -41300m

The position after 100s = S₃₀+S₇₀ = -8700 -41300 = -50 km

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Given, the mass of the body = 0.4 kg, Initial speed = 10m/s, Initial force, Retarding force, F = - 8 N (a) At t = -5s, Due to constant speed, the acceleration of the body = 0 From the relation, s = ut + (1/2)at2, we gets = 10 <math><mo>×</mo><mo> (</mo><mo>-</mo><mn>5</mn><mo>)</mo></math> + 0 <math><mo>×</mo></math> <math><msup><mrow><mrow><mo> (</mo><mfrac><mrow><mrow><mn>1</mn></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></mfrac><mo>)</mo><mo> (</mo><mo>-</mo><mn>5</mn><mo>)</mo></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></math> = -50 mAt t = 25s, The acceleration of the body due to retarding force, from the relation F = ma, a = -8/0.4 m/s2 = - 20 m/s2From the relation, s = ut + at2, , we gets = 10 <math><mo>×</mo><mo> (</mo><mn>25</mn><mo>)</mo></math> + <math><msup><mrow><mrow><mfenced separators="|"><mrow><mrow><mfrac><mrow><mrow><mn>1</mn></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></mfrac></mrow></mrow></mfenced><mo> (</mo><mo>-</mo><mn>20</mn><mo>)</mo><mo> (</mo><mo>-</mo><mn>25</mn><mo>)</mo></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></math> = -6000mAt t = 100sFor the first 30 s, while the body moves towards North, the opposite force of 8 N acts on it and from the relation s = ut + (1/2)at2, , we getS30 = 10 <math><mo>×</mo></math> 30 + <math><mfenced separators="|"><mrow><mrow><mfrac><mrow><mrow><mn>1</mn></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></mfrac></mrow></mrow></mfenced><mo> (</mo><mo>-</mo><mn>20</mn><mo>)</mo><msup><mrow><mrow><mn>30</mn></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></math> = -8700mThe speed after 30 sec, v = u+atv = 10 – (20 <math><mo>×</mo><mn>30</mn><mo>)</mo></math> = -590 m/sThe distance covered in next 70 s, S70 = -590 <math><mo>×</mo><mn>70</mn><mo>+</mo><mfenced separators="|"><mrow><mrow><mfrac><mrow><mrow><mn>1</mn></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></mfrac></mrow></mrow></mfenced><mfenced separators="|"><mrow><mrow><mn>* (0) (</mn></mrow></mrow></mfenced><mfenced separators="|"><mrow><mrow><mn>70)</mn></mrow></mrow></mfenced><mo>×</mo><mo> (</mo><mn>70</mn><mo>)</mo></math> = -41300mThe position after 100s = S30 +  S70 = -8700 -41300 = -50 km

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