A 5 kg object slows from +6 m/s to +2 m/s. What is the impulse?

Physics

1 answer:

Liula [17]3 years ago

5 0

Impulse on the object: -20 kg m/s

Explanation:

The impulse exerted on an object is a vector quantity equal to the product between the force exerted on the object and the time interval during which the force is applied.

The impulse exerted on an object is also equal to its change in momentum. Mathematically,

$I=\Delta p = m(v-u)$

where:

m is the mass of the object

u is the initial velocity

v is the final velocity

For the object in this problem, we have

m = 5 kg

u = +6 m/s

v = +2 m/s

Therefore, its impulse is:

$I=(5)(+2-(+6))=-20 kg m/s$

The negative sign means the direction of the impulse is opposite to the direction of motion of the object.

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A builder drops a brick from a height of 15 m above the ground. The gravitational field strength g is 10 N/ kg. What is the spee

Basile [38]

The speed of the brick dropped by the builder as it hits the ground is 17.32m/s.

Given the data in the question;

Since the brick was initially at rest before it was dropped,

Initial Velocity; $u = 0$

Height from which it has dropped; $h = 15m$

Gravitational field strength; $g = 10N/kg = 10 \frac{kg.m/s^2}{kg} = 10m/s^2$

Final speed of brick as it hits the ground; $v = \ ?$

<h3>Velocity</h3>

velocity is simply the same as the speed at which a particle or object moves. It is the rate of change of position of an object or particle with respect to time. As expressed in the Third Equation of Motion:

$v^2 = u^2 + 2gh$

Where v is final velocity, u is initial velocity, h is its height or distance from ground and g is gravitational field strength.

To determine the speed of the brick as it hits the ground, we substitute our giving values into the expression above.

$v^2 = u^2 + 2gh\\\\v^2 = 0 + ( 2\ *\ 10m/s^2\ *\ 15m)\\\\v^2 = 300m^2/s^2\\\\v = \sqrt{300m^2/s^2}\\ \\v = 17.32m/s$