A brick of mass 1.5 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 1.75 m away? Ignore friction.
m/s ?
(b) Redo part (a) if the coefficient of kinetic friction is 0.09.
m/s ?
Also on this one I don't know how to begin am lost! Please help thnks
Wb = mg = 1.5kg * 9.8N/kg = 14.7 N. =
Wt. of brick.
Fb = 14.7N. @ 30 Deg.=Force of brick. Fp = 14.7*sin30 = 7.35 N. = Force parallel to roof.
Fv = 14.7*cos30 = 12.7 N. = Force perpendicular to roof.
a. h = 1.75*sin30 = 0.875 m.
Vf^2 = Vo^2 + 2g*h.
Vf^2 = 0 + 19.8*0.875 = 17.15.
Vf = 4.14 m/s. = Final velocity.
b. Same as (a).
Well, it seems like you're in a slippery situation! But don't worry, I'll try to help you slide through these physics problems with a touch of humor.
(a) Since we're ignoring friction, the only force acting on the brick is its weight (mg), where m is the mass of the brick and g is the acceleration due to gravity (approximately 9.8 m/s^2).
To find the acceleration of the brick down the inclined roof, we need to use a bit of trigonometry. The component of the weight acting parallel to the roof is mg*sin(30°), and the component perpendicular to the roof is mg*cos(30°).
Using Newton's second law (F=ma), we can equate the component of the weight parallel to the roof to the product of the mass and acceleration. So we have:
mg*sin(30°) = ma
We can now solve for the acceleration (a), which is the rate of change of velocity:
a = g*sin(30°)
Since the brick starts from rest, its initial velocity (u) is zero. We can then use the kinematic equation:
v^2 = u^2 + 2as
where v is the final velocity and s is the displacement. As we only need the final velocity, we can rearrange the equation to solve for v:
v = sqrt(2as)
Plugging in the values, we have:
v = sqrt(2*g*sin(30°)*s)
where g = 9.8 m/s^2 and s = 1.75 m. Now let's calculate the value of v:
v = sqrt(2*9.8*sin(30°)*1.75) ≈ 6.72 m/s
Therefore, the brick will be moving at approximately 6.72 m/s when it reaches the edge of the roof.
(b) Now let's redo it, but this time, we'll consider the coefficient of kinetic friction (μ = 0.09). With friction in play, an additional force will act on the brick as it slides down the roof.
The frictional force (Ff) can be calculated using:
Ff = μN
where N is the normal force exerted on the brick. In this case, the normal force is equal to mg*cos(30°).
So now we have:
Ff = μmg*cos(30°)
The net force acting on the brick, when friction is included, is:
Fnet = mg*sin(30°) - Ff
And using Newton's second law again:
Fnet = ma
We can now solve for the acceleration (a):
a = (g*sin(30°) - μ*g*cos(30°))
Now, let's go back to the kinematic equation:
v^2 = u^2 + 2as
Since the brick starts from rest, its initial velocity (u) is zero. We can then solve for v:
v = sqrt(2as)
Plugging in the values:
v = sqrt(2*(g*sin(30°) - μ*g*cos(30°))*s)
Substituting g = 9.8 m/s^2, μ = 0.09, and s = 1.75 m, we can calculate the value of v:
v = sqrt(2*(9.8*sin(30°) - 0.09*9.8*cos(30°))*1.75) ≈ 6.01 m/s
Therefore, when you include friction, the brick will be moving at approximately 6.01 m/s when it reaches the edge of the roof.
I hope that helps! Just remember, sometimes physics problems can be a bit slippery, but with a little humor and perseverance, you can slide through them.
To solve this problem, we can use the principles of Newtonian mechanics, specifically the laws of motion and the concept of force. I'll guide you through the steps to find the solution to each part of the problem:
(a) If there is no friction, only the force of gravity acts on the brick as it slides down the icy roof. We can resolve this force into two components: one parallel to the incline (mg sinθ) and one perpendicular to the incline (mg cosθ). Here, θ represents the angle of inclination (30.0°) and g is the acceleration due to gravity (9.8 m/s^2).
To find the acceleration of the brick as it slides down the roof, we need to consider the component of gravitational force parallel to the incline. We can calculate the acceleration using the formula:
a = g sinθ
Substituting the given values:
a = (9.8 m/s^2) x sin(30.0°)
a ≈ 4.9 m/s^2
Now, to find the final velocity of the brick at the edge of the roof, we can use the following equation of motion:
v^2 = u^2 + 2as
Where:
v = final velocity of the brick (which we need to find)
u = initial velocity of the brick (0 m/s, as it starts from rest)
a = acceleration of the brick (4.9 m/s^2)
s = displacement (1.75 m)
Rearranging the equation, we get:
v^2 = 0 + 2(4.9 m/s^2)(1.75 m) = 17.15
Taking the square root of both sides, we find:
v ≈ √(17.15) ≈ 4.1 m/s
Therefore, when the brick reaches the edge of the roof, its speed will be approximately 4.1 m/s.
(b) In this case, we need to take into account the coefficient of kinetic friction (μ = 0.09), which opposes the motion of the brick.
The force of friction can be calculated using the formula:
frictional force = μ × normal force
The normal force is the component of the gravitational force perpendicular to the incline (mg cosθ). Thus, the frictional force is given by:
frictional force = μ × (mg cosθ)
The net force acting on the brick, now considering friction, can be calculated as:
net force = mg sinθ - frictional force
Using Newton's second law of motion (F = ma), we can set up this equation of motion for the brick moving down the roof:
net force = ma
Substituting the expressions for the net force and frictional force:
mg sinθ - μ × (mg cosθ) = ma
Rearranging the equation and solving for a:
a = g (sinθ - μ cosθ)
Now, we can use the same equation of motion as in part (a) to find the final velocity of the brick.
v^2 = u^2 + 2as
Substituting u = 0 m/s, a = g (sinθ - μ cosθ), and s = 1.75 m:
v^2 = 0 + 2g(s sinθ - μ s cosθ)
Simplifying further:
v^2 = 2g(s sinθ - μ s cosθ)
Taking the square root of both sides, we find:
v = √[2g(s sinθ - μ s cosθ)]
Substituting g ≈ 9.8 m/s^2, θ = 30.0°, μ = 0.09, and s = 1.75 m, we can calculate v.
v ≈ √[2 × 9.8 m/s^2 × (1.75 m × sin(30.0°) - 0.09 × 1.75 m × cos(30.0°))]
After calculating the expression, the value of v will be the final velocity of the brick at the edge of the roof with the frictional force taken into account.
I hope this explanation helps! Let me know if you have any further questions or need clarification.