Question

Suppose the rocket is coming in for a vertical landing at the surface of the earth. The captain adjusts the engine thrust so that rocket slows down at the rate of 2.05 m/s2 . A 6.50-kg instrument is hanging by a vertical wire inside a space ship.Find the force that the wire exerts on the instrument.

Answer 1

Answer:

R= 78.32 N

Explanation:

Given that

Acceleration ,a= 2.05 m/s²

Mass , m = 6.5 kg

The force due to acceleration

F= mass x Acceleration

F=  ma

F= 6.5 x 2.05 N

F= 13.32 N

The force due to weight

F' = m g

F' = 6.5 x 10 N             ( take g= 10 m/s²)

F'= 65 N

Therefore the net total force will be summation of force due to weight and force due to acceleration

R= F + F'

R= 65 + 13.32 N

R= 78.32 N

Answer 2

The force that the wire exerts on the instrument is R= 78.32 N

Calculation of force:

Given that

• Acceleration ,a= 2.05 m/s²

• Mass , m = 6.5 kg

• The force due to acceleration

Since The force due to weight

$F= ma\\\\F= 6.5 \times 2.05 N\\\\F= 13.32 N$

Now

$F' = m g\\\\F' = 6.5 \times 10 N ( take\ g= 10 m/s^2)$

F'= 65 N

Now

R= F + F'

R= 65 + 13.32 N

R= 78.32 N

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