A 6.00 kg friction less glider, attached to the end of an ideal spring with forc

Question

A 6.00 kg friction less glider, attached to the end of an ideal spring with force constant k = 900 N/m. undergoes simple harmonic motion with an amplitude 0.300 m. 'Start the clock' so x=0.00m at t=0.00sec and with initial motion to the right Compute the maximum speed of the glider m/s Compute the first time when the glider is at x = - 0.100 m (note negative sign). seconds A carefully designed experiment can measure the gravitational force between a pair of 3.00 kg masses. Given that the density of gold is 11,000 kg/m^3. what is the (very expensive experiment) gravitational force between two gold spheres that are touching? N You are going to lift that John Deere 4020 tractor (mass = 4000 kg) over your head (2.00 m vertical displacement). Calculate the work required to do this. You will very carefully (at the center of mass) and slowly (no tossing of the tractor allowed!) lift. If you want, you can use a pulley system (If' ideal' this does not change the energy required to lift the tractor, but it definitely reduces the force required to do so.) J How many McRib Sandwiches (500 Calories per sandwich according to the McDonald's website) must you metabolize to supply the energy for this feat? (Note this is food, so 'watch' the Calorie stuff). Assume the body converts 100% of the food energy into mechanical energy. (Bad assumption but it makes the calculation much easier.) Sandwiches

Explanation / Answer

Q1.

mass=m=6 kg

spring constant=k=900 N/m

then angular frequency=w=sqrt(k/m)=12.2474 rad/s

amplitude=A=0.3 m

then position at any time t is written as

x(t)=A*sin(w*t)=0.3*sin(12.2474*t) m

part a:

speed of the glider=dx/dt=0.3*12.2474*cos(12.2474*t) m/s=3.67422*cos(12.2474*t) m/s

as maximum value of cosine function is 1,

maximum value of speed=3.67422 m/s

part b:

let at time t, x(t)=-0.1 m

==>0.3*sin(12.2474*t)=-0.1

==>sin(12.2474*t)=-0.3333

==>12.2474*t=3.4814 seconds

==>t=0.284256 seconds

hence at t=0.284256 seconds, the glider at x=-0.1 m

Q2:

volume of the gol sphere=mass/density=3/11000=2.72727*10^(-4) m^3

if radius of the sphere is r,

then (4/3)*pi*r^3=2.72727*10^(-4)

==>r=0.04023 m

distance between two spheres=2*radius=2*0.04023=0.08046 m

hence gravitational force=G*3*3/0.08046^2=9.2783*10^(-4) N

Q3.work required=potential energy gained by the tractor

=mass*acceleration due to gravity*height

=4000*9.8*2=78400 J

b)1 cal=4.184 J

hence calories required=78400/4.184=18738 cal

hence number of sandwitches=18738/500=37.476

hence you need to eat 38 sandwitches

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---