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A 1.5-kg object has a velocity of 5j m/s at t = 0. It is accelerated at a constant rate for five seconds after which it has a velocity of (6i + 12j ) m/s. What is the magnitude of the resultant force acting on the object during this time interval?

Respuesta :

Answer:

2.76 N

Explanation:

m = mass of the object = 1.5 kg

v₀ = initial velocity at t = 0, = 0 i + 5 j

v = final velocity of the object at t = 5, = 6 i + 12 j

t = time interval = 5 sec

a = acceleration of the object = ?

Acceleration of the object is given as

[tex]a = \frac{v - v_{o}}{t}[/tex]

inserting the values

a = ((6 i + 12 j) - (0 i + 5 j))/5

a = (6 i + 7 j)/5

a = 1.2 i + 1.4 j

magnitude of the acceleration is given as

|a| = √((1.2)² + (1.4)²)

|a| = 1.84 m/s²

magnitude of the resultant force is given as

|F| = m |a|

|F| = (1.5) (1.84)

|F| = 2.76 N