Please help with my understanding of this Physics lab Work and Kinetic Energy La
ID: 1897471 • Letter: P
Question
Please help with my understanding of this Physics lab
Work and Kinetic Energy Lab, investigating the relationship between the work done on a cart and the change in the kinetic energy of the cart. We used a cart, track, 3- 20 g masses, and a photogate.
Basically we place started with the 20 g mass attached to the end of a string. We moved the cart to the end of the track so the 20 g mass was hanging at 80 cm above the floor. We then measured distance from the bottom on the 20 g mass to the floor and recorded it below and release the cart. We did this again for each mass. Below is the data recorded.
Mass HeightWork Velocity Total Kinetic Energy
20g 42cm? .507cm/s ?
40g42cm ?.691 cm/s?
60g42cm ? .876 cm/s?
Can you help me understand how to calculate the work done by the falling mass? W=mgh The car and the picket fence used had a total mass of 500g.
1)Calculate the work done by the falling mass in the table above.
2)Calculate kinetic energy of cart and the falling mass. *The car and the picket fence used had a total mass of 500g.
3) Finally when comparing the work done by the falling mass to the total change in kinetic energy for the cart and the falling mass, why might there be a difference?
Explanation / Answer
Consider cart and falling mass as one system. The only external force on the system is gravity. Now the work done by gravity (W= m_{fall}gh) where, (m_{fall} = falling mass) From the data h = 38 cm in each case. # The First Reading: Work = 20 x 980 x 38 erg = 744800 erg Total Kinetic Energy (cart + falling mass) = 66.83274 erg (delta E = Work - Total Kinetic Energy = 744733.16726 ) # The Second Reading: Work = 40 x 980 x 38 erg = 1489600 erg Total Kinetic Energy (cart + falling mass) = 128.91987 erg 1489471.08013 2.0000063722286164048613665876063 # The Third Reading: Work = 60 x 980 x 38 erg = 2234400 erg Total Kinetic Energy (cart + falling mass) = 214.86528 erg 2234185.13472 2.9999807084461501038586092312399 We clearly see that work is not equal to the total kinetic energy thus seemingly defying the famous energy conservation principle. But we know that the friction has not been accounted for. So we need to check if the differences in work and total kinetic energy doubles* (or triples) when we double (or triple) the falling mass before claiming the invalidity of the Conservation priciple. let (delta E = Work - Total Kinetic Energy) First case: (delta E_1 = 744733.16726 erg) Second Case: (delta E_2 = 1489471.08013 erg) ( rac{delta E_2}{delta E_1} = 2.0000063722286164048613665876063 pprox 2) Third case: (delta E_3 = 2234185.13472 erg) ( rac{delta E_3}{delta E_1} = 2.9999807084461501038586092312399 pprox 3) So there you go!! Friction is doing the required work to reduce the kinetic energy.