A brick of mass 0.7 kg slides down an icy roof inclined at 30.0° with respect to the horizontal.
(a) If the brick starts from rest, how fast is it moving when it reaches the edge of the roof 2.50 m away? Ignore friction
rwq
To find the velocity of the brick when it reaches the edge of the roof, we can use the principles of Newtonian mechanics.
First, let's break down the forces acting on the brick. Since there is no friction mentioned in the problem, the only forces acting on the brick are its weight (mg) and the normal force (N) exerted by the roof surface on the brick.
The weight of the brick can be calculated using the formula:
Weight (W) = mass (m) x acceleration due to gravity (g)
W = 0.7 kg x 9.8 m/s^2 = 6.86 N
The normal force (N) is perpendicular to the roof's surface and can be found by using the component of the weight perpendicular to the roof. This component can be calculated using trigonometry:
Component of weight perpendicular to the roof (Wp) = W sin θ
Wp = 6.86 N x sin 30° = 3.43 N
Since the only force responsible for the acceleration is the component of weight parallel to the roof, we can write:
Net Force = Component of weight parallel to the roof
So, let's calculate the net force:
Net Force = ma
The acceleration (a) can be calculated using the equation:
a = g sin θ
a = 9.8 m/s^2 x sin 30° = 4.9 m/s^2
Substituting the values into the equation for net force:
Net Force = 0.7 kg x 4.9 m/s^2
Net Force = 3.43 N
Now that we know the net force, we can calculate the acceleration:
Net Force = ma
3.43 N = 0.7 kg x a
a = 3.43 N / 0.7 kg = 4.9 m/s^2
With the acceleration, we can now use the kinematic equation to find the final velocity (v) of the brick when it reaches the edge of the roof:
v^2 = u^2 + 2as
Since the brick starts from rest, the initial velocity (u) is 0:
v^2 = 0^2 + 2 x 4.9 m/s^2 x 2.5 m
v^2 = 24.5 m^2/s^2 x 2.5 m
v^2 = 61.25 m^2/s^2
Taking the square root of both sides:
v = √61.25 m^2/s^2
v ≈ 7.83 m/s
Therefore, the brick will be moving at approximately 7.83 m/s when it reaches the edge of the roof.