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Two constant forces act on an object of mass m = 5.60 kg object moving in the xy plane as shown in the figure below. Force F1 is 27.5 N at 35.0°, and force F2 is 41.0 N at 150°. At time t = 0, the object is at the origin and has velocity (4.70i + 2.55j) m/s 150° 35.0 (a) Express the two forces in unit-vector notation (b) Find the total force exerted on the object. (c) Find the object's acceleration m/s Now, consider the instant t = 3.00 s (d) Find the object's velocity m/s (e) Find its position (f) Find its kinetic energy from ½mvf2 (g) Find its kinetic energy from ½mvi2 ΣF·

Respuesta :

Answer:

Part a)

[tex]F_1 = 22.5\hat i + 15.7 \hat j[/tex]

[tex]F_2 = -35.5 \hat i + 20.5 \hat j[/tex]

Part b)

[tex]F = -10 \hat i + 36.2 \hat j[/tex]

Part c)

[tex]a = -1.78 \hat i + 6.46 \hat j[/tex]

Part d)

[tex]v_f = -0.64 \hat i + 21.93\hat j[/tex]

Part e)

[tex]r_f = 6.09\hat i + 36.72 \hat j[/tex]

Part f)

[tex]KE = 1347.7 J[/tex]

Part g)

[tex]KE = 1348 J[/tex]

Explanation:

Part a)

first force is given as

[tex]F_1 = 27.5 N[/tex] at 35 degree

[tex]F_1 = 27.5 cos35 \hat i + 27.5 sin35 \hat j[/tex]

[tex]F_1 = 22.5\hat i + 15.7 \hat j[/tex]

Second force is given as

[tex]F_2 = 41 N[/tex] at 150 degree

[tex]F_2 = 41 cos150 \hat i + 41 sin150\hat j[/tex]

[tex]F_2 = -35.5 \hat i + 20.5 \hat j[/tex]

Part b)

Total force is given as

[tex]F = F_1 + F_2[/tex]

[tex]F = (22.5 - 35.5)\hat i + (15.7 + 20.5)\hat j[/tex]

[tex]F = -10 \hat i + 36.2 \hat j[/tex]

Part c)

as we know

F = ma so object acceleration is given as

[tex]a = \frac{F}{m}[/tex]

[tex]a = \frac{-10}{5.60} \hat i + \frac{36.2}{5.60} \hat j[/tex]

[tex]a = -1.78 \hat i + 6.46 \hat j[/tex]

Part d)

By kinematics we know that

[tex]v_f = v_i + at[/tex]

[tex]v_f = (4.70 \hat i + 2.55 \hat j) + (-1.78\hat i + 6.46\hat j)(3)[/tex]

[tex]v_f = -0.64 \hat i + 21.93\hat j[/tex]

Part e)

As we know that final position is given as

[tex]r_f = v_i t + \frac{1}{2}at^2[/tex]

[tex]r_f = (4.70 \hat i + 2.55 \hat j)(3) + \frac{1}{2}(-1.78 \hat i + 6.46 \hat j)(3^2)[/tex]

[tex]r_f = 6.09\hat i + 36.72 \hat j[/tex]

Part f)

final kinetic energy is given as

[tex]KE = \frac{1]{2}mv_f^2[/tex]

[tex]KE = \frac{1}{2}(5.60)(0.64^2 + 21.93^2)[/tex]

[tex]KE = 1347.7 J[/tex]

Part g)

final kinetic energy is given as

[tex]KE = KE_i + F.r[/tex]

[tex]KE = \frac{1}{2}(5.60)(4.70^2 + 2.55^2) + (-10 \hat i + 36.2 \hat j).(6.09\hat i + 36.72\hat j)[/tex]

[tex]KE =80 + 1268.4[/tex]

[tex]KE = 1348 J[/tex]