Besides the gravitational force, a 2.80-kg object is subjected to one other constant force. The objectstarts from rest and in 1.20 s experiences a displacement of (4.20 i - 3.30 j) m, where the direction of jis the upward vertical direction. Determine the other force.

the other force= (16.3i + 14.6j)N

EXPLANATION:

Given:

Mass=2.80-kg

t= 1.2s

Since the object started from rest, the origin is (0,0) which symbolize the the object's initial position.

We will need to calculate the magnitude of the displacement using the below formula;

d = (1/2)at2 + v0t + d0

But note that

d0 = 0,( initial position)

v0 = 0( initial position)

a is the net acceleration

d = √[4.202 + (-3.30)2] m = 5.34 m

Hence, the magnitude of the displacement is 5.34 m, then we can make 'a' the subject of formula in the above expression in order to calculate the value for acceleration, note that d0 = 0,( initial position) and v0 = 0( initial position)

d = (1/2)at2

a = 2d/t2 = 2(5.34)/(1.20)2 m/s2 = 7.42 m/s2

the net acceleration is 7.42 m/s2

Acceleration in terms of the vector can be calculated as

a=2(ri - r0)/t^2

Where t =1.2s which is the time

a= 2(4.2i - 3.30j)/ 1.2^2

a=( 5.83i - 4.58j)m/s

now the net force can now be calculated since we have known the value of acceleration, using the formula below;

F(x) = ma - mg

Where a = 5.83i - 4.58j)m/s and g= 9.8m/s

2.8(5.83i - 4.58j)m/s - (2.80 × 9.8)m/s^2

Therefore, the other force= (16.3i + 14.6j)N

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