The change in the total energy of the object is zero (0).
The given parameters:
work done by the machine, W = 50 J
mass of the object, m = 10 kg
To find:
the change in the total energy of the object
The change in the total energy of the object is the difference between the objects initial energy due to its position and the work done on the object.
Based on work energy-theory, the work done on the object is equal to the energy of the object.
- The energy of the object = work-done on the object
- The change in total energy = 50 J - 50 J = 0
Thus, the change in the total energy of the object is zero (0).
Learn more here: brainly.com/question/20377140
Greater the mass greater is inertia. Greater the inertia greater is the force required to stop motion of an object. Linear motion depends only on mass whereas rotational motion depends on mass, size and shape of an object. So conclusion is that it would be difficult to stop 10 kg mass cuz of greater inertia compared to 1kg mass. Hope it clears your doubt.
Solution
distance travelled by Chris
\Delta t=\frac{1}{3600}hr.
X_{c}= [(\frac{21+0}{2})+(\frac{33+21}{2})+(\frac{55+47}{2})+(\frac{63+55}{2})+(\frac{70+63}{2})+(\frac{76+70}{2})+(\frac{82+76}{2})+(\frac{87+82}{2})+(\frac{91+87}{2})]\times\frac{1}{3600}
=\frac{579.5}{3600}=0.161miles
Kelly,
\Delta t=\frac{1}{3600}hr.
X_{k}=[(\frac{24+0}{2})+(\frac{3+24}{2})+(\frac{55+39}{2})+(\frac{62+55}{2})+(\frac{71+62}{2})+(\frac{79+71}{2})+(\frac{85+79}{2})+(\frac{85+92}{2})+(\frac{99+92}{2})+(\frac{103+99}{2})]\times\frac{1}{3600}
=\frac{657.5}{3600}
\Delta X=X_{k}-X_{C}=0.021miles
