a) You are driving across a gravel bed, when you see a patch of ice ahead. You slam on the brakes attempting to stop before you hit the ice. The kinetic energy dissipated and the braking maneuver was 3.0 X 10^5J. The maneuver brings your vehicle (assume the vehicle mass is 1.0 X 10^3 kg) to rest at the edge of the ice patch. What was your initial velocity?
b) In attempting to turn your car around, you inadvertently end up stuck on the patch of ice (assume it is a frictionless surface) and have to be pulled off the surface by a tow truck with a cable and winch. The two truck cable is at an angle 40.0 degrees with respect to the horizontal surface. The tension in the cable during the pulling process is 3.50 X 10^3 N. The car has a mass of 1.0 X 10^3 kg. How much work is done by the cable in moving the car a distance of 15 m?
c) The tow truck driver pulls your vehicle onto the flat-bed. What is the change in grav. potential energy of the car after it is raised 1.8 m above.Physics 3 part question, this is a little long...?A) The car has kinetic energy embodied by its relative speed. The brakes dissipate this energy as a frictive loss, largely by turning it into heat and by ablation, until the car has no kinetic energy left, and has stopped.
Kinetic energy is: KE = (1/2)mv^2
B) Only force acting the same direction as the motion produces work. The force of tension from the truck's cable can be divided into two components by:
Fx = Tcos40, Fy = Tsin40
The mass of the object moved is irrelevent, work is proportional to the force exerted and the distance moved by:
W = Fd
C) As the truck is raised, it gains potential gravitational energy. If it were suddenly dropped from a height, it would release energy in the form of friction, sound (essentially friction with the air), and possibly a deformation of the vehicle amd/or ground. Gravitational potential energy is proportional to the objects mass, it's height above the surface, and the gravitational constant, g (g ~= 9.81ms^-2 or 32.0 fts^-2) by:
PE = mghPhysics 3 part question, this is a little long...?1) KE = 1/2*m*v^2 =%26gt; v = sqrt(2*KE/m) = ...
2) W = F*d*cos@ = ...
3) PE = m*g*h = ...
