Question

A 400 g particle moving along the z-axis experiences the force shown in (Figure 1). The particle's velocity is 2.0m/s at x = 0m. You may want to review (Pages 214 - 218) 

Part A 

What is its velocity at x= 3 m? 

Express your answer to two significant figures and include the appropriate units. 

Answer 1

ANSWER :

Mass of particle = 400 g = 0.4 kg

Force applied = 15 N from x = 0 to x = 1.

Now,

F = m * a 

=> a = F/m = 15/0.4 = 37.5 m/sec^2 .

Let  t be the time taken to move from x = 0 to x = 1.

So,

As per formula : s = u t + 1/2 a t^2

=> 1 = 2.0 t + 1/2 (37.5) * t^2 

=> 18.75 t^2 + 2.0 t - 1 = 0

=> t = - 2/ (2*18.75)  +/- sqrt(2^2 - 4*18.75*(-1)) / (2*18.75)

=> t = - 0.29 , 0.1837 secs.

Discard the negative value which is meaningless.

=> t = 0.1837 sec.

Hence, velocity at x = 1, V1 = u + a t = 2 + 37.5 * 0.1837 = 8.89 m/sec

From x = 1 to x = 3, F is decreasing at a constant rate. So, we can take average F as acting during this interval.

Hence, F’ = (15 + 0)/2 = 7.5 N in the interval (1, 3).

So, acceleration a’ in that interval = F’/a’ = 7.5/0.4 = 18.75 m/sec^2

For moving from x = 1 to x = 3, if time taken is t’,

then :

s = V1 t’ + 1/2 a’ t’^2

=> 2 = 8.89 * t’ + 1/2 * 18.75 t’^2 

=> 9.375 t’^2 + 8.89 t’ - 2 = 0

=> t’ = - 8.89/(2*9.375) + sqrt(8.89^2 - 4*9.375*(-2)) / (2*9.375)

(We have discarded the negative value of t’)

=> t ‘ = 0.1878 sec.

So final velocity at x = 3, V3 

= V1 + a’ t’ 

=> V3 = 8.89 + 18.75 * 0.1878

=> V3 = 12.41 m/sec.

So, velocity at x = 3, is 12.41 m/sec (ANSWER).

Answer 2

mass of particle 400 g 3 0 a m = 0.400 kg Velocity Vi = 2.0m/s Fx (N) 15 when alm) Work done by force when particle is moving - on to d = 3m J F da 0 3.3 between area under fan curve X om and x3m W = (1-0) *( 15-0) + 4x( 3-1 * 15 15 & 15 - 30T

kinehh energy W = change m ve- £m vi 11 W 2W м - ) VY 2W + T NG + 2.02 2 r 30 0.400 Т. 7. if 12mls 7 f Velocity at x = 3m