A brick of mass 1.0kg slides down an icy roof inclined at 30degrees with respect

Question

A brick of mass 1.0kg slides down an icy roof inclined at 30degrees 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 2.0 M away? (ignore friction) (Hint: USE CONSERVATION OF ENERGY)

(B) Now consider the problem if friction acts on the brick. SKETCH A FREE BODY DIAGRAM SHOWING A COORDINATE SYSTEM AND THE FORCES ACTING ON THE BRICK. (Hint: you prbably want to make one of the rectangular axes parallel to the direction of motion)

(C) If the brick starts from rest, how fast is it moving when it reaches the edge of the roof 2.0M away if the coefficient of friction is 0.10? (Hint: Friction does negative work on the brick, slowing it down. FIND THE WORK DONE BY FRICTION. Then USE THE WORK ENERGY THEROEM TO FIND SPEED OF BRICK.)

Explanation / Answer

Conservation of energy states that Kinetic Energy + Potential energy = Total energy at any and all points.

At rest, Kinetic Energy = 0 and Potential energy = mgh.

m = mass

g = acceleration due to gravity(9.81m/s^2)

h = height relative to another fixed point, ie: the end of the 2m roof, or 2*sin(30)

Kinetic energy is equal to .5mv^2

m = mass

v = velocity

Since we are assuming Kinetic energy at the start to equal zero and Potential energy at the end to equal zero, then Potential energy at the start = Kinetic energy at the end.

From this we obtain: (1kg)*(9.81m/s^2)*(2m)*sin(30) = .5*(1kg)*(v)^2

Through algebra we obvain v to equal 4.41 m/s

Now for part C:

Since the normal force due to gravity on the object is F = m*a, or Force = mass times gravity, we obtain F = 9.81N.

Since the 9.81N is directed perpendicular to an angled slope, we multiple it by sin(theta), or sin(30) in this case.

Thus the normal force that resists the acceleration due to gravity.

Now, we have the equation for the force due to friction as 9.81N*sin(30)*.1 = .49N

Since the work energy theorm states the energy of a system at one point is equal to the energy of a system minus the work done by the system at another point, we obtain: (1kg)*(9.81m/s^2)*(2)*sin(30) = .5(1kg)*(v)^2-.49N

From this we obtain the answer to part C is: 4.317m/s

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