Two constant forces act on an object of mass m = 5.00 kg moving in the xy plane

Question

Two constant forces act on an object of mass m = 5.00 kg moving in the xy plane as shown in Figure P7.61. Force is 25.0 N at 35.0 degree, and force F _2 is 42.0 N at 150 degree. At time t = 0, the object is at the origin and has velocity (4.00i + 2.50j ) m/s, Express the two forces in unit-vector notation. Use unit-vector notation for your other answers. Find the total force exerted on the object. Find the object's acceleration. Now, considering the instant t = 3.00 s, find the object's velocity, its position, its kinetic energy from and its kinetic What conclusion can you draw by comparing the answers to parts (f) and (g)?

Explanation / Answer

mass m=5 kg

force F1=25 N at 35 degrees

F2=42 N at 150 degrees

a)

force F1=(Fx)i+(Fy)j

F1=F1*cos(theta)i+F1*sin(theta)j

F1=25*cos(35)i+25*sin(35)j

F1=(20.48)i+(14.34)j in N

and

force F2=(Fx)i+(Fy)j

F2=F2*cos(theta)i+F2*sin(theta)j

F2=42*cos(150)i+42*sin(150)j

F2=(-36.37)i+(21)j in N

b)

total force Fnet=F1+F2

Fnet=(20.48)i+(14.34)j + (-36.37)i+(21)j

Fnet=(-15.89)i+(35.34)j in N

c)

acceleration a=Fnet/m

a=(-15.89)i+(35.34)j/5

a=(-3.18)i+7.07)j m/sec^2

d)

at t1=0 sec, v1=4i+2.5j

at t2=3 sec, v2=?

a=v2-v1/(t2-t1)

==>

v2=v1+a(t2-t1)

v2=(4i+2.5j)+(-3.18)i+7.07)j*(3-0)

v2=(4i+2.5j)+(-9.54)i+21.21)j

v2=(-5.54)i+(23.71)j in m/sec

e)

x=v1*t+1/2*a*t^2

x=(4i+2.5j)*3+1/2*(-3.18)i+7.07)j*3^2

x=(-10.31)i+(34.31)j in m

f)

K.E=1/2*m*v2^2

magnitude of v2=sqrt(-5.54^2+23.71^)=23.05 m/se

K.E=1/2*5*(23.05)^2

K.E=1328.25 J

g)

K.E=1/2*m*v1^2+F.dr

=1/2*5*(4^2+2.5^2)+((-15.89)i+(35.34)j.(-10.31)i+(34.31)j)

=55.62+163.82+1212.51

=1431.95 J

the k.E in part f is not equals to part g

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