A 1200-kg automobile travels at 90.0 km/h. What net work would be required to bring it to a stop?

Answer 1

By the Work-Energy theorem the net work-done by all the forces on an object is equal to the change in its kinetic energy.

#W_{Tot}=\DeltaK=K_f-K_i=-375\quad kJ#

Work-Energy Theorem: The net work-done by all the forces acting on a body is equal to the change in its kinetic energy.

#W_{Net}=\Delta K=K_f-K_i#

Since the object comes to rest finally, #K_f=0#.

If #v_0# is the initial speed of the object and #m# its mass, then #K_i=1/2mv_0^2#

Therefore, #W_{Net}=\DeltaK=K_f-K_i=0-1/2mv_o^2=-1/2mv_0^2#

#v_0 = 90 \quad (km)/(hr) = 25\quad m/s; \qquad m = 1200\quad kg#

#W_{Net} = - 3.75\times10^5\quad J = -375 \quad kJ#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.