Ask question

Ask question

All categories

3 months ago

14

The magnitude of the force associated with the gravitational field is constant and has a value FF . A particle is launched from

point BB with an initial velocity and reaches point AA having gained U0U0 joules of kinetic energy. A resistive force field is now set up such that it is directed opposite the gravitational field with a force of constant magnitude 12F12F . A particle is again launched from point BB . How much kinetic energy will the particle gain as it moves from point BB to point AA ?

1 answer:

serg [3.5K]3 months ago

7 0

Answer:

The particle's kinetic energy will equal 12U₀.

Explanation:

Given that,

A particle is fired from point B with an initial speed and arrives at point A with U₀ joules of kinetic energy gained.

The consistent force acting is 12F.

As per the problem,

The kinetic energy is

$U_{0}=Fx$ ....(I)

Constant force remains at 12F.

A resistive force field now exists,

With the resistive force defined as,

$F_{r}=12F$

As the particle travels from point B to point A,

We need to find the kinetic energy

Using the kinetic energy formula

$U=F_{r}x$

Substituting values for $F_{r}$

$U=12Fx$

Then, from equation (I)

$U=12U_{0}$

Consequently, the particle’s kinetic energy will amount to 12U₀.

You might be interested in

It's a snowy day and you're pulling a friend along a level road on a sled. You've both been taking physics, so she asks what you

Maru [3345]

Answer:

0.0984

Explanation:

The first diagram below illustrates a free body diagram that will aid in resolving this problem.

According to the diagram, the force's horizontal component can be expressed as:

$F_X = F_{cos \ \theta}$

Substituting 42° for θ and 87.0° for $F$

$F_X =87.0 \ N \ *cos \ 42 ^\circ$

$F_X =64.65 \ N$

Meanwhile, the vertical component is:

$F_Y = Fsin \ \theta$

Again substituting 42° for θ and 87.0° for $F$

$F_Y =87.0 \ N \ *sin \ 42 ^\circ$

$F_Y =58.21 \ N$

In resolving the vector, let A denote the components in mutually perpendicular directions.

The magnitudes of both components are illustrated in the second diagram provided and can be represented as A cos θ and A sin θ

The frictional force can be expressed as:

$f = \mu \ N$

Where;

$\mu$ is the coefficient of friction

N = the normal force

Also, the normal reaction (N) is calculated as mg - F sin θ

Substituting $F_Y \ for \ F_{sin \ \theta}$ . Normal reaction becomes:

$N = mg \ - \ F_Y$

By balancing the forces, the horizontal component of the force equals the frictional force.

The horizontal component is described as follows:

$F_X = \mu \ ( mg - \ F_Y)$

Rearranging the equation above to isolate $\mu$ leads to:

$\mu \ = \ \frac{F_X}{mg - F_Y}$

Substituting in the following values:

$F_X \ = \ 64.65 \ N$

m = 73 kg

g = 9.8 m/s²

$F_Y = \ 58.21 N$

Thus:

$\mu \ = \ \frac{64.65 N}{(73.0 kg)(9.8m/s^2) - (58.21 \ N)}$

$\mu = 0.0984$

Therefore, the coefficient of friction is = 0.0984

5 0

3 months ago

A trained sea lion slides from rest down a long

Keith_Richards [3271]

Answer:

1.5 m/s²

Explanation:

Begin by sketching a free body diagram. Three forces are at play on the sea lion: the force of gravity acting downwards, the normal force that is perpendicular to the ramp, and the frictional force parallel to the ramp.

Considering the forces perpendicular to the incline:

∑F = ma

N − mg cos θ = 0

This gives us N = mg cos θ

Next, examining the forces parallel to the incline:

∑F = ma

mg sin θ − Nμ = ma

Substituting for N yields:

mg sin θ − (mg cos θ) μ = ma

g sin θ − g cos θ μ = a

hence a = g (sin θ − μ cos θ)

If we set θ = 23° and μ = 0.26:

a = 9.8 (sin 23 − 0.26 cos 23)

this results in a = 1.48

When rounded to two significant figures, the acceleration of the sea lion is 1.5 m/s².

5 0

1 month ago

A center lane with solid and broken yellow lines that is used by vehicles making left turns in both directions is called a:

Sav [3153]

A left turn lane is designated for vehicles making left turns. Vehicles must completely enter this lane and wait for a clear path before proceeding with the turn. It is not permissible for a vehicle to be partially in and out of the lane, as this obstructs traffic and creates hazardous conditions.

7 0

2 months ago

A window washer on a hanging platform cleans the outside of windows on a skyscraper. The washer hoists the platform up the side

kicyunya [3294]

Answer:

The acceleration of the platform is - 1.8 m/s²

Explanation:

The net force on a body causes that body to accelerate in the direction of the resultant force applied.

Setting up the force equilibrium for the configuration:

ma = 800 - mg

100a = 800 - 100×9.8

100a = - 180

100a = - 180

a = - 1.8 m/s²

This indicates that the body is falling downward.

6 0

3 months ago