a) Work done = Net Kinetic Energy
= 1/2 x 50 kg x ((12m/s)^2 - (3m/s)^2)
= 0.5 x 50 Kg x (144 -9)(m/s)^2
= 3375 Kg (m/s)^2
b) Force = mxa
a = 120 N/50 Kg = 2.4 m/s^2
Using newtons third law of motion, we get-
V^2 - U^2 = 2 x a x S
S= (12^2-3^2)m^2/s^2/(2 x 2.4 m/s^2)
= 28.125 m
Answer:
![\boxed {\boxed {\sf 6 \ meters \ per \ second}}](https://tex.z-dn.net/?f=%5Cboxed%20%7B%5Cboxed%20%7B%5Csf%206%20%5C%20meters%20%5C%20per%20%5C%20second%7D%7D)
Explanation:
Speed can be found by dividing the distance by the time.
![s=\frac{d}{t}](https://tex.z-dn.net/?f=s%3D%5Cfrac%7Bd%7D%7Bt%7D)
The distance is 18 meters and the time is 3 seconds.
![d= 18 \ m \\t= 3 \ s\\](https://tex.z-dn.net/?f=d%3D%2018%20%5C%20m%20%5C%5Ct%3D%203%20%5C%20s%5C%5C)
Substitute the values into the formula.
![s=\frac{18 \ m }{ 3 \ s }](https://tex.z-dn.net/?f=s%3D%5Cfrac%7B18%20%5C%20m%20%7D%7B%203%20%5C%20s%20%7D)
Divide.
![s= 6 \ m/s](https://tex.z-dn.net/?f=s%3D%206%20%5C%20m%2Fs)
The speed of the puck is <u>6 meters per second.</u>
B. It contains more matter