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Three forces acting on an object are given by F1 = (−2.00 i + 2.00 j) N,F2 = (5.00 i − 3.00 j) N, and F3 = (−45.0 i) N. The object experiences an acceleration of magnitude 3.75 m/s2.

i. What is the direction of the acceleration?

ii. What is the mass of the object?

iii. If the object is initially at rest, what is its speed after 10.0 s?

iv. What are the velocity components of the object after 10.0 s?

Respuesta :

From summation and resultant of forces, the four answers are:

i.  ∅ = 34 degrees

ii. M = 0.48 kg

iii. V = 37.5 m/s

iv. V = 37.5i + 0j + 0k

Given that Three forces acting on an object are

F1 = (−2.00 i + 2.00 j) N

F2 = (5.00 i − 3.00 j) N

F3 = (−45.0 i) N.

If the object experiences an acceleration of magnitude 3.75 m/s2.

i. The direction of the force will be the direction of the acceleration.

Sum of the forces = (-2 + 5 - 4.5)i + (2 - 3 + 0)j

Sum of the forces = -1.5i -j

The acceleration direction will be

tan∅ = [tex]\frac{y}{x}[/tex]

tan∅ = [tex]\frac{1}{1.5}[/tex]

∅ = [tex]tan^{-1}[/tex] (0.67)

∅ = 34 degrees (approximately)

ii. The mass of the object can be calculated from Newton's law.

Resultant force = mass x acceleration.

Resultant force = [tex]\sqrt{1.5^{2} + 1^{2} }[/tex]

Resultant force = [tex]\sqrt{3.25}[/tex]

Resultant force = 1.8N

Then,

1.8 = 3.75M

M = 1.8 / 3.75

M = 0.48 kg

iii. If the object is initially at rest, the speed of the object after 10.0 will be calculated by using first equation of motion.

V = U + at

Where U = 0

V = 0 + 3.75 x 10

V = 37.5 m/s

Therefore, the speed of the object after 10.0 is 37.5 m/s

iv. The velocity components of the object after 10.0 are

V = 37.5i + 0j + 0k

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