Question

A 5 Kg bucket is being lifted by Sue straight up. A)If Sue is lifting the bucket up with constant velocity with what force is she lifting the bucket with? B) If Sue uses the same force and lifted the bucket on the moon which has a gravitational pull of 1.6 m/s2, with what acceleration will the bucket rise?

Answer 1

Answer:

A) Sue is lifting the bucket by a force of 49.035 newtons.

B) The bucket has an acceleration of 8.207 meters per square second on the Moon.

Explanation:

A) According to the First Newton's Law, a system is at equilibrium when it is either at rest or travelling at constant velocity. In this case, Sue must exert an external force on the bucket, whose magnitude is equal to the weight of the bucket but direction is opposed to it. By Second Newton's Law, we find that:

$\Sigma F = F - m\cdot g = 0$ (1)

Where:

$F$ - Lifting force, measured in newtons.

$m$ - Mass of the bucket, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

If we know that $m = 5\,kg$ and $g = 9.807\,\frac{m}{s^{2}}$, then the lifting force is:

$F = m\cdot g$

$F = (5\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$F = 49.035\,N$

Sue is lifting the bucket by a force of 49.035 newtons.

B) By the Second Newton's Law, we have the following model:

$\Sigma F = F-m\cdot g = m\cdot a$ (2)

Where $a$ is the net acceleration of the bucket, measured in meters per square second.

If we know that $F = 49.035\,N$, $m = 5\,kg$ and $g = 1.6\,\frac{m}{s^{2}}$, then the net acceleration of the bucket is:

$a = \frac{F}{m} -g$

$a = \frac{49.035\,N}{5\,kg}-1.6\,\frac{m}{s^{2}}$

$a = 8.207\,\frac{m}{s^{2}}$

The bucket has an acceleration of 8.207 meters per square second on the Moon.

Answer 2

(a) The force applied by Sue in lifting the bucket at a constant velocity is 49 N.

(b) The acceleration of the bucket when lifted on the moon is 8.2 m/s².

The given parameters;

• mass of the bucket, m = 5 kg

The force applied by Sue in lifting the bucket at a constant velocity is calculated as;

$F = m(a + g)$

• at constant velocity, a = 0

$F= mg\\\\F = 5 \times 9.8\\\\F = 49 \ N$

The acceleration of the bucket when lifted on the moon with the calculated force is;

$F = m(a + g)\\\\a+g = \frac{F}{m} \\\\a = \frac{F}{m} - g\\\\a = \frac{49}{5} - 1.6\\\\a = 8.2 \ m/s^2$

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