Besides the gravitational force, a 2.30-kg object is subjected to one other cons

Question

Besides the gravitational force, a 2.30-kg object is subjected to one other constant force. The object starts from rest and in 1.20 s experiences a displacement of (4.45î 3.30) m, where the direction of is the upward vertical direction. Determine the other force. (Express your answer in vector form.)

Desides the gravitational torce, a 2.30-kq object is sub ected to one other constant torce. The object starts trom rest and in 1.20 s experiences a displacement ot (4.45i 3.30j) m, where the direction of j is the upward vertical direction, Determine the other torce Express your answer in vactor form.)

Explanation / Answer

Place the origin (0,0) at the initial position of the object. The magnitude of the displacement is

d = (1/2)at2 + v0t + d0

d0 = 0,v0 = 0

a = net acceleration

d = sqrt[4.452 + (-3.30)2] m = 5.54 m

d = (1/2)at2

a = 2d/t2

a = 2(5.54)/(1.20)2 m/s2

a = 7.69 m/s2

F = ma = net force

f = unknown constant force

F = f - mgj

f = F + mgj

Let be the angle from the +x axis to the displacement vector

= tan-1(-3.30/4.45) = 323.44°

F = Fxi + Fyj

F = ma cos i + ma sin j

f = ma cos i + m(a sin + g) j

f = m[a cos i + (a sin + g) j]

f = 2.30 [7.69 cos323.44° i + (7.69 sin323.44° + 9.81) j]

f = (14.21 i + 12.03 j) N

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