Volume 5. Physics 208 A.
last update on 8 May, 1998
1. Accelerate a 1000 kg mass to 5m/s on a frictionless surface. The force to carry out this task could be small or big. There is no lower threshold of force. If you apply a tiny force, the acceleration is small but continued application will obtain the desired velocity.
a = F/m. v = at = F/m * t.
F=v*m/t = 5*1000/t. The longer the time, the smaller the necessary force.
2. For a F = 500 N, a = F/m = 500/1000 = 0.5 m/s/s
3. With a very small force. a = F/m = 2/1000 = .002 m/s/s
4. A skate boarder falls and slides to a halt in 5m having an initial velocity of 14m/s.
v^2 = 2ax. a = v^2/(2x) = 14^2/(2*5) = 19.6 m/s/s
The force is F = ma = 70*19.6 = 13720 Newtons.
1. When you are standing on the floor, why aren’t you accelerated downwards? After all, your weight is a force which is pulling you in the downward direction. Identify all the forces.
Your weight on the chair causes a reaction from the chair on your body which is upwards. The net force on you is the weight down and reaction up. These are equal and opposite so the net force is zero and so acceleration is zero.
2. for a leap of 1 meter, a 70 kg athlete accelerates herself upwards by pushing down on the floor for a distance of 0.2 meters.
a) What is the takeoff speed.?
Falling one meter obtains a speed of v^2 = 2gh. v=sqrt(2*10*1) = 4.5m/s The jump is just the time reverse of this so the launch speed must be 4.5m/s upwards.
b) what is the acceleration that is required to achieve this take-off speed.
The student obtains the speed of 4.5m/s within a distance of 0.2m so a=v^2/2d = 4.5^2/(2*.2) = 50 m/s^2 or 5 g. This is a rather large acceleration.
c) How much net upward force must be obtained by pressing on the floor?
The net upward force must be = ma = 70*50 = 3500N , a rather large force.
3. The cyclist is going at a constant velocity so the net force must be zero. The wheel pushes backward on the ground with 25 N. The reaction force pushes the bike inthe forward direction with 25 N. This is counteracted by the backward push of the air resistance which must be also 25 N since the net force is zero.
The only forces on the free body diagram are: weight down, ground reaction up; wind resistance backwards , and reaction force of ground in the forward direction.
4. An 50 kg athlete accelerates up to 10m/sec. How much kinetic energy is acquired?
K = 1/2 mv^2 = 1/2 * 50 * 10^2 = 2500J
If this occurs in 3 seconds,
a)how much power was required?
power = energy/time = 2500J/3 sec = 833 watts.
b)What is the average acceleration?
a = delv/delt = 10/3 = 3.3m/sec/sec
c) Is this power the major part of what an athlete must produce for sprinting? Explain.
The major power that the sprinter expends is in the acceleration of the limbs in the cycling of the running motion. A sprinter uses a lot higher power than just the 833 watts
1. An athlete sheds some fat and jumps higher. The original jump used energy mgh while the new jump has energy'= m'gh'. In this case, energy = energy'. So mgh=m'gh', so
h' = mh/m' = 80*0.6/70 = 1.14*0.6 = 0.69m, a considerable amount!
2. With superb neurological training, the athlete can produce 50% more force. this amounts to 50% more energy. Work = F*d= mgh. Work' = F'*d=mgh'. Divde one by the other to get F'*d/F*d=mgh'/mgh. h'/h = F'/F=1.5. You can jump 50% higher!!
1. continuation of B-2.
d) Is the force produced by your hips and legs equal to (c)? greater? lesser? Explain and calculate what the net force should be.
The net force is composed of the upward force from the floor and the weight pushing down. net force = f - w. f = net force + w = 3500+700= 4200N. or 6 x your weight.!!
e) What is the power required for this jump?
The average power = F * avg speed = 4200 * 4.5/2 =9450 watts. You need power to jump high!!
2. An 50 kg athlete accelerates up to 10m/sec. How much kinetic energy is acquired?
K = 1/2 mv^2 = 1/2 * 50 * 10^2 = 2500J
If this occurs in 3 seconds,
a)how much power was required?
power = energy/time = 2500J/3 sec = 833 watts.
b)What is the average acceleration?
a = delv/delt = 10/3 = 3.3m/sec/sec
c) Is this power the major part of what an athlete must produce for sprinting? Explain.
The major power that the sprinter expends is in the acceleration of the limbs in the cycling of the running motion. A sprinter uses a lot higher power than just the 833 watts
Questions on Conservation of Energy.
1. A pole vaulter approaches the pit with a velocity of 10m/s.
a)With a perfectly elastic performance, how high will the vaulter
raise his center of mass?
K=1/2mv^2 = mgh
h = 1/2 v^2/g = .5*10^2/10 = 5m
b) If the vaulter loses 20% of his energy, because of technique, how high can he raise his center of mass?
mgh = .8 K = .8 1/2 mv^2
h = .8*(height at 100% efficiency) = 4 m
2. Where does the rest of energy go if only 25% of the food energy intake of an athlete goes toward work.
3. A 1.3 kg sphere is dropped through a tall column of liquid.
When the sphere has fallen a distance of 0.5m, it is observed to
have a velocity of 2.5m/s. [First you should identify all the
energy sources and where the energy can go.]
a) How much work was done by the frictional viscosity of the
liquid?
The ball has gravitational potential energy, kinetic energy and energy lost to friction.
After it has fallen .5m, the change in grav pot. energy is mgh = 1.3 * 10 *.5 =6.5J
The kinetic energy is K = 1/2 mv^2 = 1/2 * 1.3 * 2.5^2 = 4 J
This means that 2.5 J has gone to stir up the liquid.
b) What was the average force of friction?
W = F*d ; F = W/d = 2.5J/.5m = 5 newtons
c) If the ball were to be doubled in radius [but without changing it's mass] describe quantitatively the change in passage of the ball. eg it's velocity after falling .5m; how much energy is lost to the liquid.
Ff = 1/2 r v^2 Cd * A.
the Frictional force would be 4 X larger because of the increase in area of the ball.
The ball would have a much smaller velocity at the .5m depth.
d) What was the approximate radius of the ball in the original problem.
In water, r = 1000 kg/m^3, Cd = .5.
A = 2Ff/(v^2Cdr) = 2*5 /(1.25^2 * .5 * 1000) = .018m^2.
radius = .05m.