Question

10. What are the and the resulting What are the the velocity components of the mass after 4.251 seconds of acceleration v magnitude 1d. What are the magnitude and direction of the acceleration of the mass after 4.251 seconds of acceleration? v direction le. What are the x and y coordinates of the mass after 4.251 seconds of acceleration? R 1f. What are the magnitude and direction of the displacement vector R of the mass after 4.251 seconds of acceleration?

Answer 1

resultant Force

F = 24.46 i +241.1 j ( +x points to E and +y to N)

mass m = 12.25 kg

acceleration:

a_x = F_x/m = 24.46/12.25 = 2.0 m/s/s

a_y = F_y/m = 241.1 /12.25 = 19.68 m/s/s

Velocity after t = 4.251 s

v_x = a_x*t   = 2.0 *4.251 = 8.502 m/s

v_y = a_y*t = 19.68* 4.251 = 83.67 m/s

acceleration magnitude a = sqrt ( 2.0² + 19.68²) = 19.78 m/s/s

direction : tan(p) = a_y/a_x = 19.68/2.0

p = 84.2^o NoE  

direction of magnitude of acceleration are const. do not change with time

Vel. mag. at 4.251 s = sqrt ( 8.5² + 83.67²) = 84.1 m/s

direction Tan(p) = 83.67/8.5

It will be same as a i.e. 84.2 NoE , does not change with time

position:

X = 0.5a_xt² = 0.5*2.0*4.251² = 18.07 m

Y = 0.5*19.68 *4.251² = 177.8 m

displacement vector:

R = 18.07 i + 177.8 j  

magnitude |R| = sqrt ( 18.07² + 177.8²) = 178.72 m

direction same as a and v i.e. tan^-1 ( 177.8/18.07)

= 84.2^o NoE , will not change with time