When your sled starts down from the top of a hill, it hits a frictionless ice slick that extends all the way down the hill. At the bottom, the ground is dry and level. The effective coefficient of friction between the sled runners and the ground is 0.62. If the hill is 50 m high, how far will your sled travel once it reaches the bottom

Explanation:

Given that,

Height of hill is 50m

Coefficient of friction is μ=0.62

Energy is conserved, then

K.E at the bottom of the hill is equally to P.E at the top of the hill

½mv²=mgh

Mass cancel out

½v²=gh

v²=2gh

v=√2gh

Since g=9.81 and h=50

v=√2×9.81 ×50

v=31.32m/s

This is the initial speed at the bottom of the hill

At the bottom of the hill 3 forces are acting on the body

1. Weight,

2. Normal

3. Frictional force

Now taking newton law of motion

ΣF = ma.

Along y axis, since the body is not moving in y direction, they acceleration along y is 0m/s²

ΣFy = 0

N-W=0

N=W

Since weight =mg

N=W=mg

Using law of friction, Fr=μN

Therefore,

Fr=μmg

Applying Newton law to the x-direction

ΣFx = ma

Fr=ma,

Since Fr=μ mg

μ mg=ma

a=μg

Given that μ=0.62 and g=9.81

a=0.62×9.82

a=6.076m/s²

Since we have acceleration, we can use any of the equation of motion to find distance travel

v²=u²-2gS, . this is due that the box is decelerating and it comes to a halt and this show that final velocity v=0m/s

Then,

0²=31.32²-2×9.81 ×S

-31.32² =-2×9.81×S

S=31.32²/(2×9.81)

S=50.05m

The sled will travel a distance of 50.1m once it reach the bottom

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west