Tammi+Nichols

__ **Week four (13 th February, 2012)** __ 1. A 0.5 kg rock is projected from the edge of the top of a building with an initial velocity of 10.5 m/s at an angle 54º above the horizontal. Due to gravity, the rock strikes the ground at a horizontal distance of 15.3 m from the base of the building. Assume: The ground is level and that the side of the building is vertical. The acceleration of gravity is 9.8 m/s². What is the height of the building

First step is to list all the givens, which will make it easier to see what you will need. m=0.5 kg Vo= 10.5 m/s at 54 degrees with the horizontal (X-Xo)= 15.3 m (Y-Yo)=?

Next we break down the initial velocity into x and y components. Vix= 10.5cos(54)= 6.17 m/s Viy= 10.5sin(54) = 8.49 m/s

Then acceleration x and y components Ax= 0 Ay= 9.8 m/s^2

Now we use the formula (X-Xo)= Vo*t+1/2Axt^2 since Ax is 0 the 1/2Axt^2 cancels out leaving: (15.3m)=.6.17m/s(t) therefore t=15.3m/6.17m/s t=2.48s

Now that we've found t we can solve for (Y-Yo) (Y-Yo)=Voyt+1/2Ayt^2 (Y-Yo)=(8.49m/s)(2.48s)+1/2(-9.81m/s^2)(2.48s)^2 (Y-Yo)=-9.17 m

the answer is negative, only because we chose upwards as our positive direction. The displacement of the rock downward will actually be the absolute value of the number we found.

Professor's Note : Good job ! final answers are not necessary --- keep that in mind for the next week. __ **Week 08 (March, 12th, 2012)** __

01. A loaded sled weighing 85 N rests on a plane inclined at an angle of 20⁰ with the horizontal. The coefficents of static and kinetic friction between the inclined plane and the sled are 0.25 and 0.15 respectively. a). What is the least magnitude of the force F, parallel to the plane, that will prevent the sled from slipping down the plane b). What is the minimum force required to start the sled moving upward? c). What value of F (force) is required to move the sled up the plane at constant velocity?

a). Since the sled is not moving, acceleration equals 0. We can assume that the normal force is equal to the y component of the weight. The friction force is towards the same direction as the force F. a=0 Fn=Wsin20 The sum of Fx= F+fs-wcos20=0 F=wcos20-fs the least magnitude of F is the x component of weight minus the friction force, or the normal force times the coefficient of static friction.

b). Since the sled is not yet moving, but about to, the acceleration is still 0. The static friction force is working against the Force. Fn=Wsin20 a=0 sum of all Fx= F-fs-Wcos20=o F=fs+Wcos20 F=Fn*.25+85cos20 The minimum force to start the sled moving is equal to the normal force times the coefficient of static friction plus the x component of the weight.

c). The velocity of the sled is constant, therefore the acceleration of the sled is again 0. a=0 This time the friction force is kinetic. sum of all Fx=F-fk-Wcos20=0 F=fk+Wcos20 F=Fn*.15+85cos20 The Force necessary to move the sled at a constant velocity is equal to the normal force times the coefficient of kinetic friction plus the x component of the weight.