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A 11.50 kg object has the given and acceleration components.

=(0.67ms2)+(0.81ms3)
=(11.7ms2)−(0.63ms3)
What is the magnitude net of the net force acting on the object at time =4.47 s ?

User Marimaf
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1 Answer

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Answer:

To find the net force acting on the object, we need to add the force components along each axis. We are given the acceleration components, which we can use to find the force components using Newton's second law:

F = ma

Along the x-axis:

F_x = ma_x = (11.7 kg)(0.67 m/s^2) = 7.839 N

Along the y-axis:

F_y = ma_y = (11.7 kg)(-0.63 m/s^2) = -7.371 N

The net force is the vector sum of the x and y components of force:

F_net = sqrt(F_x^2 + F_y^2) = sqrt((7.839 N)^2 + (-7.371 N)^2) = 10.925 N

Therefore, the magnitude of the net force acting on the object at time t = 4.47 s is 10.925 N.

User Luiz Martins
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