Question

A 3900 kg car is parked at the top of an 8.7 m long driveway that is sloped at 17? with the horizontal. The parking brake fails.

If an average friction force of 4200 N im- pedes the motion, find the speed of the car at the bottom of the driveway. The acceleration of gravity is 9.8 m/s2 .

Starting from rest, a 10.0 kg suitcase slides 3.00 m down a frictionless ramp inclined at 31.0? from the floor. The suitcase then slides an additional 5.00 m along the floor before coming to a stop.

The acceleration of gravity is 9.81 m/s2 .

a) Find the speed of the suitcase at the bottom of the ramp.

Answer in units of m/s

b) Find the coefficient of kinetic friction be- tween the suitcase and the floor.

c) Find the change in mechanical energy due to friction.

An 102 kg skydiver jumps out of an airplane at an altitude of 1508 m and opens the parachute at an altitude of 143 m. The total retarding force on the diver is constant at 50 N with the parachute closed and constant at 2708 N with the parachute open.

The acceleration of gravity is 9.8 m/s2 .

What is the speed of the diver when he lands on the ground?

Answer in units of m/s

b) At what height should the parachute be opened so that the final speed of the sky- diver when he hits the ground is 4 m/s?

Answer in units of m

Answer 1

problem 1:

the horizontal component of weight acts towards the bottom of the inlcine.

W(h)=Wcos17

W(h)=36549.97 N

friction acts in the opposite direction.

acceleration=[W(h)-friction]/mass

=8.29 m/s^2

v^2=u^2-2*a*S

v=    \(\sqrt{2*a*S}\)   

v=12.01 m/s

problem 2:

decrease in potential energy of the suitcase=increase in its kinetic energy

m*g*lsin31=0.5*m*v^2

therefore, v=    $+\(\sqrt{2*g*lsin31}\)+$   

normal force acting on the suitcase when it is on the floor=m*g

kinetic friction=u*normal force=u*m*g

decrease in kinetic energy=work done to oppose friction

0.5*m*v^2=u*m*g*s where, u=coefficient of kinetic friction

s=distance travelled on the floor

therefore,

u=(0.5*v^2)/(g*s)

b)u=0.31

mechanical energy of the suitcase=kinetic energy+potential energy

when the suitcase is on the floor, it has only kinetic energy

c)change in mechanical energy=work done by friction

=-u*m*g*h work done by friction is negative because friction force acts opposite to the displacement of the suitcase

=-152.055

problem 3:

increase in kinetic energy of the diver=decrease in potential energy-work done by retarding force

0.5*m*v^2=m*g*h-F(closed)*h1-F(open)*h2

F(closed)=retarding force acting on the diver with parachute closed=50 N

F(open)=retarding force acting on the diver with parachute open=2708 N

h=1508 m

h1=1508-143=1365 m

h2=143 m

on solving, you will get,

a)v=143.62 m/s

let the parachute be opened at height x.

increase in kinetic energy of the diver=decrease in potential energy-work done by retarding force

0.5*m*4^2=m*g*h-F(closed)*(1508-x)-F(open)*x

816=1431996.8+50x-2708x

2658x=1431180.8

b) x=538.44 m

Answer 2