Physics help PLEASE?

Question

Starting from rest, a child on a sled slides 20m down an icy hill that is inclined at 28 degrees. The coefficient of kinetic friction for the sled on ice is .09. After the child slides down the hill, they slide on level ground on gravel with a coefficient of kinetic friction of .33. The child and the sled together are 18kg. How far from the base of the hill does the child on the sled slide before coming to rest? Any help would be appreciated!!!

Answer 1

This will be solved using conservation of energy.
Since friction does work that removes energy from the system, the equations are


sliding down the hill from rest

KE=m*g*sin(28)*20
-m*g*cos(28)*µ(ice)*20


Where KE is the kinetic energy at the bottom of the slope
the slide across the gravel converts the KE into heat due to friction

KE=m*g*µ(gravel)*d

Since the KE is the same in both

m*g*µ(gravel)*d
=m*g*SIN(28)*20
-m*g*COS(28)*µ(ice)*20


Simplify

µ(gravel)*d=SIN(28)*20
-COS(28)*µ(ice)*20

d=20*(SIN(28)
-COS(28)*µ(ice))/µ(gravel)

=23.6 m

j

Answer 2

First find vf at the bottom of the hill.

vf^2 - vi^2 = 2ax

where vi is zero and x is 20m
a is the acceleration or force over mass
forces: weight minus friction
weight is mg sin(28)
friction is mg cos(28)time the coefficient on ice

Once you have that velocity, use the same equation again, but the vf you solved for becomes vi and the vf becomes zero.

Now the force is just the friction mg times the coefficient on gravel.

Solve that equation for x.

Good luck!

Answer 3

Work done by friction = friction force * distance
friction force = normal force * coefficient of friction

Thus, work done by friction = normal force * coeff. of friction * distance

Remember, the normal force is always perpendicular to the surface, and gravity always acts downward.

On the hill, the work done by friction (Wf) is:
Wf = m*g*cos(28)*friction_coeff*d
Wf = 18*9.81*cos(28)*0.09
Wf = 14.0 Joules

The potential Energy at the top of the hill is equal to
PE = m*g*h
PE = m*g*20*sin(28)
PE = 18*9.81*20*sin(28)
PE = 1,658.0 Joules

The Energy at the bottom of the hill is equal to PE - Wf
PE - Wf = 1658 - 14 = 1644 Joules of kinetic energy.

Since the sled is sliding to a stop, need to find the distance "d" where the work done by friciton will be equal to the remaining 1644 Joules of energy

Wf_flat = m*g*friction_coeff_gravel*d

1644 = 18*9.81*0.33*d

Solving for d yields:
d = 28.2 metres.

Answer 4

btw Its a very small child!

Do this in two parts:

The hill:

Draw a diagram

Find R = 18*9.81/cos28 = 200N

Fr = 0.09*200 = 18N

Resolving along the plane:

18*9.81*cos28 - 18 = 18a

a = 7.66ms^-2

v = as = 7.66*20 = 153m/s

At the base:

a = -0.33*9.81 = -3.24ms^-2 (they are slowing)

s = s , u=153, a=-3.24, v=0

v^2 = u^2 + 2as

0 = 23409 - 2*3.24*s

s = 23409/6.48 = 3.6km

ok so I've made a mistake here =|