Answer:
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
