A 10.0-kg object is initially moving with a velocity of 20.0 m/s to the north and is acted on by a constant net force. after the

Question

gtnhenbr [62]

3 years ago

8

A 10.0-kg object is initially moving with a velocity of 20.0 m/s to the north and is acted on by a constant net force. after the

object moves 30.0 m to the north, its velocity is 12.0 m/s north. what is the constant net force acting on the object?

Physics

2 answers:

ValentinkaMS [17] · Answer 1 · 2022-06-10T07:18:25+03:00

The direction of motion of the object did not change, so we can find the force that acted on the object by using the work-energy theorem, which states that the work done by the force is equal to the variation of kinetic energy of the object:
$W=K_f - K_i$
$Fd= \frac{1}{2}mv_f^2- \frac{1}{2}mv_i^2$
where
F is the force
d is the distance through which the force has been applied (30.0 m)
$m=10.0 kg$ is the object mass
$v_i=20.0 m/s$ is the initial velocity of the object
$v_f=12.0 m/s$ is its final velocity

Re-arranging the formula, we find the magnitude of the force:
$F= \frac{m(v_f^2-v_i^2)}{2d}= \frac{(10.0 kg)((12m/s)^2-(20m/s)^2)}{2 \cdot 30.0 m}=-42.7 N$
And the negative sign means the force is in the opposite direction of the motion of the object (in fact, the object is decelerating)

Ksju [112] · Answer 2 · 2022-06-10T08:59:14+03:00

The constant net force acting on the object is about 42.7 N to the south.

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<h3>Further explanation</h3>

Newton's second law of motion states that the resultant force applied to an object is directly proportional to the mass and acceleration of the object.

$\large {\boxed {F = ma }$