Mass and Kinetic Energy Essay

The goal of the beam project is to design and construct a beam that can hold a given amount of weight without breaking. The beam is required to hold a concentrated load of 375 lbf on the X-axis and 150 lbf on the Y-axis. The maximum allowable weight of the beam is 250 grams. The maximum allowable deflection for the beam is 0.230 in. and 0.200 in. for the X and Y-axis respectively. The beam is required to be 24 in. in length, and it will be tested on a simply supported configuration spanning 21 in. All calculations are to be done under the assumption that the density of basswood is 28 lbm/ft3 and the modulus of elasticity for basswood is 1.46x106 lbm/in2. Given the constraints of a spending cost of $10.50, a maximum beam weight of 250 grams,

A chandelier with a weight of 100 N is suspended at a single point by

3. Calculate the N component of a resultant 32.5 m/s, 35.0 º E of N.

2. The ball contains kinetic energy while in motion near the bottom of its path.

After collecting data and analyzing the results the group concluded that tension will always be greater than mass for uniform constant motion due to the presence of θ. As the length of the string for the flying toy increased, θ decreased in both trials for the cow and bat. For instance, when the bat had a length of 0.86m, the calculated θ was 34.4°, but when the length was shortened to 0.61m, the θ increased to 36.5°. The period was also shortened when the length decreased; therefore, velocity decreased as string length increased while acceleration increased with a

5. Tie one end of the 60-cm string to the mass. Place the mass on a table below

12. Forces of 3 N and 4 N act at right angles on a block of mass 5 kg.show that the resulting acceleration is 1 m/?

Mass of the Fulcrum Clamp: _________________(grams) Mass of Clamp with V-wire: ________________(grams) Mass of Weight Hanger: __________________(grams) Balancing point (center of mass) of meter stick, x0 = ____________(cm) Values Case 1 m1 = 100 g m2 = 200 g x1 = 15 cm x2 =_________ Moment (lever) arms r1 = ___________ r2 = ___________ Results *(see note above) τcc = ___________ τcw = ___________ % Diff. __________ Case 2 m1 = 100 g m2 = 200 g m3 = 50 g

After doing this problem, I found that I am very comfortable using the kinematic equation of vf = v​i​​+at, when I use this equation I find that I always get the right answer for the problem, which is great, and I have no further questions to help me understand this

The boat sails forward into a weighted ball that has gravitational potential energy. The boat passes its momentum to the weighted ball at the top of a ramp. The ball falls down onto one end of a catapult. Once the weighted ball falls down onto one side of the catapult, the other side with a non weighted ball is driven up launches the ball up into a tube. The ball travels through the short tube, and passes its momentum to another ball at the top of a ramp. The ball has gravitational potential energy, and it falls down the ramp losing that energy turning it into kinetic energy. That ball hits a peg on a wheel. This wheel has four pegs on it, and one of them has a smaller peg perpendicular to the original peg. The wheel is turned driving the peg with the smaller peg up and into a block chain. The peg transfers its energy into the first block, and the first block transfers its energy into the next and so on. The final block hits a pin which is inside a slot. A balloon is positioned behind the pin, so once the block hits it, the pin is driven forward into the balloon. The pin pokes a hole in the side of the balloon releasing the air inside. The balloon is

5. What’s really meant is that the forces are the same but going in different directions as a mirror image of one another. So when one force is going for example 2 newtons in a positive direction, the same amount is going in the opposite direction.

As a result of trail and error method we find the angle for the third pulley and the mass that should had be suspended from it. This will balance the forces deployed on the strings due to the other two masses. While the third force is defined as the equilibrant (������⃗������) Since it is the force that establishes the equilibrium. It is also the negative of the resultant -������⃗������ = ������⃗������ = ������⃗ ������ + ������⃗������. We gathered and recorded the mass and the angle required for the third pulley to enable to put the system into the equilibrium in table 1.

II) TO DETERMINE THE VALUES OF RA AND RB AS A GIVEN LOAD MOVES FROM ONE END OF A SIMPLY SUPPORTED BEAM TO THE OTHER

The beam was loaded the mid-length in 2.745 lbs. increments up to 6.745 lbs. The change in clearance of every load step was measured and data was recorded.

Table 1 & 2: First, find the mass of the wooden block and record the data. Then place the wooden block on the inclined plane (at 0o) with the wide side down. The height of the pulley should be the same height as the screw location on the wooden block. Then hang a weight on the opposite side of the hanger and add weights until block starts to move with a constant velocity (push block to overcome fs¬). Then record the resulted weight of the hanger in Table 1 (as F). Add 500 g to the wooden block and repeat the process. Replace 500 g with 1 kg on the wooden block. Repeat the process described above.