A 23.8-pound object has the following acceleration components: What is the magni

Question

A 23.8-pound object has the following acceleration components:

What is the magnitude of the net force acting on the object at time t = 4.07 s?

What is the direction of the net force at this same time? Give your answer as a number of degrees counter-clockwise from the x axis.

A 23.8-pound object has the following acceleration components Im a.-0.55 0.77 Im a,0.5 0.91 091-31t What is the magnitude of the net force acting on the object at time t4.07 s? Number Fnet 937.65 N What is the direction of the net force at this same time? Give your answer as a number of degrees counter-clockwise from the +x axis. Number e- 82.17

Explanation / Answer

  

At  t = 4.07 s

ax=(.55M/s2)+(.77m/s3)t
ax = 0.55 + 0.77x.4.07 = 3.683 m/s²

ay=(10.5m/s2)-(.91m/s3)t
ay = 10.5 - 0.91x4.07 = 6.79 m/s²

You can add these components and use F=ma, where 'a' is the resultant acceleration. or you can work out Fx and Fy and combine these. It doesn't' make any difference....

F_x = m*a_x = 23.8 x 3.683 = 87.65 N
F_y = m*a_y = 23.8 x6.796 = 161.74 N

|F| = (F_x² + F_y²)
= (87.65² + 161.74²)
= 183.96N

Angle = tan-1(F_y/F_x) = tan-1(161.74/87.65) = 61.54

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