Two forces F vector_1 = (-4.70 i cap + 3.40 j cap) N and F vector_2 = (-4.95 i c

Question

Two forces F vector_1 = (-4.70 i cap + 3.40 j cap) N and F vector_2 = (-4.95 i cap + 5.35 j cap) N, act on a particle of mass 2.10 kg is initially at rest at coordinates (-2.30 m, +4.40 m). (a) What are the components of the particle's velocity t = 10.5s ? v vector = m/s In what direction is the particle moving at t = 10.5 s ? degree counterclockwise from the +x-axis (c) What displacement does the particle undergo during the first 10.5 s? delta r vector = m What are the coordinates of the particle at t = 10.5 s? x = m y = m

Explanation / Answer

From the given question,

F1= -4.7i + 3.4j

F2= -4.95i+ 5.35j

mass=2.10 kg

initial coordinates(-2.3, 4.4)

(a) Net force acting F=F1+F2

=-4.7i + 3.4j-4.95i+ 5.35j

=-9.65i +8.75j

Net acceleration= F/m= (-9.65i +8.75j)/2.1=-4.59i +4.16j

Final velocity after 10.5 sec= (u+at)i + (u+at)j

=(-4.59 x 10.5)i + (4.16 x 10.5)j

=-48.19i + 43.68j

(b)direction of velocity= 360- tan-1(43.68/48.19)

=317.81 counter clockwise

(d)displacement of particle

=(-2.3+ut+ (1/2)at2)i + (4.4+ut+ (1/2)at2)j

=(-2.3+(1/2)(-9.65)(10.5)2 )i + (4.4+(1/2)(8.75)(10.5)2)j

=-534.25 i + 486.74j

( coordinates of particle at 10.5 sec)

(c)dispalcement of the particle=(-534.25-(-2.3))+ (486.74-4.4)=-531.95i +482.34j

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