Even with no friction, it depends on the slope of the roof. That is, it depends on how much elevation (altitude) he loses during the slide.
Whatever that number is ... call it 'h' ... Santa's speed when he reaches the edge is
Square root of (19.6h) meters per second.
It doesn't matter how much he weighs, or how far he has slud. Only how much altitude he lost on the slope while sliding.
Answer:
60 N
Explanation:
This is just Newton's Second Law
F = m*a
F = ?
m = 12 kg
a = 5 m/^2
F = 5*12 = 60 Newtons
Answer:
The object will move to Xfinal = 7.5m
Explanation:
By relating the final velocity of the object and its acceleration, I can obtain the time required to reach this velocity point:
Vf= a × t ⇒ t= (7.2 m/s) / (4.2( m/s^2)) = 1,7143 s
With the equation of the total space traveled and the previously determined time I can obtain the end point of the object on the x-axis:
Xfinal= X0 + /1/2) × a × (t^2) = 3.9m + (1/2) × 4.2( m/s^2) × ((1,7143 s) ^2) =
= 3.9m + 3.6m = 7.5m
Answer:
Force of friction, f = 751.97 N
Explanation:
it is given that,
Mass of the car, m = 1100 kg
It is parked on a 4° incline. We need to find the force of friction keeping the car from sliding down the incline.
From the attached figure, it is clear that the normal and its weight is acting on the car. f is the force of friction such that it balances the x component of its weight i.e.
f = 751.97 N
So, the force of friction on the car is 751.97 N. Hence, this is the required solution.
