All categories

3 years ago

A girl coasts down a hill on a sled, reaching

Physics

1 answer:

Simora [160]3 years ago

8 0

The sled travels a distance of 60.2 m before coming to rest.

The girl reaches the ground with a speed u. She then travels forward on the hard icy snow, which has a coefficient of kinetic friction $\mu _k$ . She comes to rest after travelling a distance s on the ground, due to the friction between the sled and the ground.

If the force of friction is f , the mass of the girl and the sled is m and the acceleration due to gravity is g, then,

$f=\mu_kmg$ .......(1)

This force exerts a decelerating force on the sled. If the deceleration of the sled is a, then,

$f=ma$ .......(2)

From equations (1) and (2),

$f=\mu_kmg=-ma\\ a=-\mu_kg$ ......(3)

Substitute 0.038 for $\mu _k$ and 9.81 m/s²for g.

$a=-\mu_kg\\ =-(0.038)(9.81m/s^2)\\ =-0.3728m/s^2$

Since the girl comes to rest, its final velocity is 0. Substitute 6.7 m/s for u and -0.3728m/s² for a in the equation of motion

$v^2=u^2+2as$

Solve for s.

$v^2=u^2+2as\\ (0m/s)^2=(6.7m/s)^2+2(-0.3728m/s^2)s\\ s=\frac{(6.7m/s)^2}{2(0.3728m/s^2)} \\ =60.2m$

Thus, the girl travels a distance of 60.2 m before coming to rest.

You might be interested in

What are the milestones of modern phyiscs?

nlexa [21]

Answer:

The articles appearing under "Milestones in Physics" will give an insight into special events or situations that have been decisive for the evolution of Physics

6 0

3 years ago

Read 2 more answers

a basketball player can leap upward .65m how long does the basketball player remain in the air use 9.81m/s²

tigry1 [53]

At the player's maximum height, their velocity is 0. Recall that

${v_f}^2-{v_i}^2=2a\Delta y$

which tells us the player's initial velocity $v_i$ is

$0^2-{v_i}^2=-2g(0.65\,\mathrm m)\implies v_i=3.6\dfrac{\rm m}{\rm s}$

The player's height at time $t$ is given by

$y=v_it-\dfrac g2t^2$

so we find their airtime to be

$0.65\,\mathrm m=\left(3.6\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2\implies t=0.36\,\mathrm s$

6 0

3 years ago

If a builder of mass 75kg climbs a vertical ladder of 25m how much energy has she gained ?

erica [24]

So,

GPE (graviational potential energy) = mass x g x height

GPE is depends on where zero height is defined. In this situation, we define h = 0 as the initial height.

$GPE = 75 \ kg*9.8 \ \frac{m}{s^2}*25 \ m$

$GPE = 18,375 \frac{kg*m^2}{s^2}$

$GPE = 18,375 \ joules(J) \ or \ 18.375 \ kilojoules(kJ)$

The builder has gained 18.375 kJ of PE.

4 0

3 years ago

A heavy rope, 60 ft long, weighs 0.7 lb/ft and hangs over the edge of a building 140 ft high. (a) How much work W is done in pul

Marina CMI [18]

Answer:

5880lb-ft of work is done

Explanation:

The length of the heavy rope is given as 60ft and the weight per length is 0.7lb/ft.

Therefore, the total weight of the heavy rope is

60×0.7 =42lb.

The work done in pulling the heavy rope to the top of the building is w = Fd.

Where

F is force is measured in pounds;42lb

d is distance through which the heavy rope is to be pulled measured in feet; 140ft

w= 42lb×140ft= 5880lb-ft

4 0

3 years ago

A car rounds a curve. The radius of curvature of the road is R, the banking angle with respect to the horizontal is θ and the co

exis [7]

Answer:

v = √[gR (sin θ - μcos θ)]

Explanation:

The free body diagram for the car is presented in the attached image to this answer.

The forces acting on the car include the weight of the car, the normal reaction of the plane on the car, the frictional force on the car and the net force on the car which is the centripetal force on the car keeping it in circular motion without slipping.

Resolving the weight into the axis parallel and perpendicular to the inclined plane,

N = mg cos θ

And the component parallel to the inclined plane that slides the body down the plane at rest = mg sin θ

Frictional force = Fr = μN = μmg cos θ

Centripetal force responsible for keeping the car in circular motion = (mv²/R)

So, a force balance in the plane parallel to the inclined plane shows that

Centripetal force = (mg sin θ - Fr) (since the car slides down the plane at rest, (mg sin θ) is greater than the frictional force)

(mv²/R) = (mg sin θ - μmg cos θ)

v² = R(g sin θ - μg cos θ)

v² = gR (sin θ - μcos θ)

v = √[gR (sin θ - μcos θ)]

Hope this Helps!!!

5 0

3 years ago