Physics Question help :)?

Question

A girl coasts down a hill on a sled, reaching level ground at the bottom with a speed of 7.0 m/s. The coefficient of kinetic friction between teh sled's runners and the hard, icy snow is 0.050, and the girl and sled together weigh 645 N. How far does the sled travel on the level ground before coming to rest?

Answer 1

This problem involves kinetic energy, force of kinetic friction, work, and conservation of energy.

Given: weight of girl + sled=645N
mass of girl +sled=645/g
=6.58N
speed of girl +sled at bottom=7m/s
coefficient of kinetic friction=0.050

Kinetic Energy, KE at the bottom of the hill=1/2mv^2
KE=1/2*(6.58)7^2
=161.2 Joules

Force of kinetic friction, f=uR, where u=coefficient of kinetic friction, and R is the force normal to the surface equal to weight of girl+sled.

f=0.050*645
=32.25N

Work=f*s, where f is the force of kinetic friction and s is the distance the sled travels. Thus

Work=32.25s

Based on principle of conservation of energy:

KE=Work
161.2=32.25s
s=161.2/32.25
=5.0m

Answer 2

The frictional force that retards the sled is

0.05 * 645 N

The acceleration due to this is

0.05 * 645 N / (645 / g)
=0.05 * 645 / ( 645 / 9.8) m/s^2
=0.05 *9.8 m / s^2
=0.49 m/s^2

The distance to stop is

= v^2 /2a = 7^2/(2*0.49) = 49/(2*0.49) = 50 m

(Did you notice we would get the same answer if we used any other value for the weight of the sled + girl?)

Answer 3

I supposed that you will need to draw a diagram to help you solve this question.
You can label the given values on the diagram and it will be easier for you to figure out how solve this question.
I only provide you with the way to solve this question but no answers would be given to you.

Answer 4

this ques need time and brain to understand so try n find out the question yourself in any refrence books dont waste ur time asking such ques. on net

Answer 5

x=vot+1/2 at^2
x = 0 + 9.8*.5*3.48