Question

A 9.30 kg mass is moving at a constant velocity of 4.00 m/s.

Answer 1

1) The magnitude of the force is 4.37 N

2) The distance covered is 17.0 m

Explanation:

1)

In order to find the force needed to bring the mass at rest, we need to find its acceleration first.

The acceleration is given by:

$a=\frac{v-u}{t}$

where

u = 4.00 m/s is the initial velocity of the mass

v = 0 is its final velocity

t = 8.47 is the time interval

Solving the equation,

$a=\frac{0-4.00}{8.47}=-0.47 m/s^2$

Now we can find the net force acting on the mass, which is given by Newton's second law:

$F=ma$

where

m = 9.30 kg is the mass

$a=-0.47 m/s^2$ is the acceleration

Substituting,

$F=(9.30)(-0.47)=-4.37 N$

So, the magnitude of the force is 4.37 N.

2)

The distance through which the mass has moved can be found by using the following suvat equation:

$s=ut+\frac{1}{2}at^2$

where

u = 4.00 m/s is the initial velocity

t = 8.47 s is the time

$a=-0.47 m/s^2$ is the acceleration

s is the distance covered

Solving for s,

$s=(4.00)(8.47)+\frac{1}{2}(-0.47)(8.47)^2=17.0 m$

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