The magnitude of the net force exerted in the x direction on a 3.20-kg particle

Question

The magnitude of the net force exerted in the x direction on a 3.20-kg particle varies in time as shown in the figure below. (a) Find the impulse of the force over the 5.00-s time interval. vector I = 12 N middot s (b) Find the final velocity the particle attains if it is originally at rest. vector v_f = m/s (c) Find its final velocity if its original velocity is -3.30 hat i m/s. vector v_f = m/s (d) Find the average force exerted on the particle for the time interval between 0 and 5.00 s. vector F_avg = N

Explanation / Answer

I = total area

I = 1/2*2*4+1*4+1/2*2*4 = 4+4+4 = 12 kg m/s

impulse = change in momentum

vf = I/m = 12/3.2 = 3.75 m/s

c) vf = I/m+v0 = 3.75-3.3 = 0.45 m/s

d) f= I/t = 12/5 = 2.4 N

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