Question

A 6 kg object is subject to three forces

F1 = 20i + 30j N

F2 = 8i - 50j N

F3 = 2i + 2j N

Find the acceleration of the object​

Answer 1

Answer:

Given,

m=7kg

F

1

=20

i

^

+30

k

^

F

2

=8

i

^

−5

j

^

The net force is equal to the sum of two force,

F

=

F

1

+

F

2

F

=20

i

^

+30

j

^

+8

i

^

−5

j

^

=28

i

^

+25

j

^

Magnitude, ∣

F

∣=

(28)

2

+(25)

2

=35N

Force, F=ma

a=

m

F

acceleration, a=

7

37.5

=5m/s

2

.

The acceleration is 5m/s

2

.

Answer 2

Answer:

Acceleration of the body subjected to three forces will be $5.83m/s^2$

Explanation:

Total force on object = $F= F_1+ F_2+F_3$

$F= (20+8+2)i+(30-50+2)j$

$F=30i-18j$

magnitude of force = $\sqrt{30^2+18^2} = \sqrt{1224}$ = 34.98

F=ma

Acceleration: a=$\frac{F}{m}=\frac{34.98}{6}=5.83m/s^2$

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