Physics, 11.11.2020 17:50 rowdycar313p0ao5k

A 4.00-kg particle moves along the x axis. Its position varies with time according to x= 5t +1 2.0t^3, where x is in meters and t is in seconds. Find: a. the kinetic energy of the particle at any time t.
b. the acceleration of the particle and the force acting on it at time t.
c. the power being delivered to the particle at time t.
d. the work done on the particle in the interval t = 0 to t =5

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Answer from: letsgetcookingblog

a) The kinetic energy of the particle at any time t is $K = 50+120\cdot t^{2}+72\cdot t^{4}$ .

b) The acceleration of the particle at time t is $a= 12\cdot t$ . The force acting on the particle at time t is $F = 48\cdot t$ .

c) The power being delivered to the particle at time t is $\dot W = 240\cdot t +288\cdot t^{3}$ .

d) The work done on the particle in the interval t = 0 to t = 5 is 48000 joules.

Explanation:

a) The kinetic energy of the particle is entirely translational, whose formula is:

$K = \frac{1}{2}\cdot m \cdot v^{2}$ (1)

Where:

- Translational kinetic energy, measured in joules.

- Mass of the particle, measured in kilograms.

- Velocity of the particle, measured in meters per second.

The velocity of the particle is the rate of change of the position of the particle in time, that is:

$v = 5+6\cdot t^{2}$ (2)

Where is the time, measured in seconds.

By substituting on (1), we have the following expression: ( $m = 4\,kg$ )

$K = 2\cdot (5+6\cdot t^{2})^{2}$

$K = 2\cdot (25+60\cdot t^{2} +36\cdot t^{4})$

$K = 50+120\cdot t^{2}+72\cdot t^{4}$

The kinetic energy of the particle at any time t is $K = 50+120\cdot t^{2}+72\cdot t^{4}$ .

b) The acceleration of the particle is the rate of change of the velocity of the particle in time, that is:

$a= 12\cdot t$ (3)

Where is the acceleration of the particle, measured in meters per square second.

The acceleration of the particle at time t is $a= 12\cdot t$ .

The force is obtained by multiplying (3) by the mass of the particle. That is to say: ( $m = 4\,kg$ )

$F = m\cdot a$ (4)

$F = 48\cdot t$

The force acting on the particle at time t is $F = 48\cdot t$ .

c) According to the Work-Energy Theorem, the change in kinetic energy of the particle equals the change in the net work done on the particle. In this case, the power is equal to the rate of change in kinetic energy.

$\dot W = \dot K$ (5)

$\dot W = \frac{d}{dt}(50+120\cdot t^{2}+72\cdot t^{4})$

$\dot W = 240\cdot t +288\cdot t^{3}$

The power being delivered to the particle at time t is $\dot W = 240\cdot t +288\cdot t^{3}$ .

d) The work done on the particle (), measured in joules, is equal to the change of the kinetic energy of the particle:

W = K(5)-K(0) (6)

$W = [50+120\cdot (5)^{2}+72\cdot (5)^{4}]-[50+120\cdot (0)^{2}+72\cdot (0)^{4}]$

$W = 48000\,J$

The work done on the particle in the interval t = 0 to t = 5 is 48000 joules.

Answer from: Quest

it's a b and

Answer from: Quest

m₁v₁ + m₂v₂ = (m₁ + m₂)vmomentum is conserved.the left-hand side of the equation represents the momentum in the system before the collision.the right-hand side represents the momentum in the system after the collisionm₁ and v₁ represents the mass and the velocity of the physics blockm₂ and v₂ represents the mass and velocity of the ball of puttyv represents the final velocity of the combined mass.the question states the two masses stick together. they combined mass has its own velocity with the body mass the sum of the two objects' massestheir combined momentum is (m₁ + m₂)vto find the magitude of (m₁ + m₂)v, we need to add the vectors on the left-hand side of the equation: m₁v₁ + m₂v₂.note the diagram. i have positioned the vectors in accordance with their directions.the magnitude of the resultant can be calculated using the pythagorean theorem since this is a right triangle situation (north and west are perpendicular)|(m₁ + m₂)v| = √[ (m₁v₁)² + (m₂v₂)² ]|(m₁ + m₂)v| = √[ (1.5kg·8.0m/s)² + (0.40kg·20m/s)² ]|(m₁ + m₂)v| = 14.4 kg m/sfor two significant figures (assuming that 20m/s is two), this rounds to 14 kg m/s

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