Please Show work and steps 1. A 68-kg trampoline artist jumps upward from the to

Question

Please Show work and steps

1.

A 68-kg trampoline artist jumps upward from the top of a platform with a vertical speed of 4.2 m/s .

PART A : How fast is he going as he lands on the trampoline, 2 m below?

PART B: If the trampoline behaves like a spring of spring constant 5×104 N/m , how far does he depress it?

2.

A ski starts from rest and slides down a 28 incline 65 m long.

PART A: If the coefficient of friction is 0.090, what is the ski's speed at the base of the incline?

PART B: If the snow is level at the foot of the incline and has the same coefficient of friction, how far will the ski travel along the level? Use energy methods.

3.

The first car has twice the mass of a second car, but only half as much kinetic energy. When both cars increase their speed by 12 m/s , they then have the same kinetic energy.

PART A: What were the original speeds of the two cars?  

Explanation / Answer

Q1.

part A:

initial veloicty=4.2 m/s

displacement=-2 m

acceleration=-9.8 m/s^2

then using the formula,

final speed^2-initial speed^2=2*acceleration*displacement

we get

final speed^2-4.2^2=2*(-9.8)*(-2)

==>final speed=sqrt(4.2^2+2*9.8*2)=7.54 m/s

part B:

let the uncompressed height of the trampoline (i.e. 2 m below the platform) be our referrence level for potential energy

i.e. at that point, potential energy =0 and height should be measured from that point.

as measured in part A, at this level, the artist's speed=7.54 m/s

as potential energy is zero, total mechanical energy=kinetic energy=0.5*mass*speed^2=0.5*68*7.54^2=1932.9544 J

lets assume that he compresses the trampoline by distance d.

at that point he will come to rest. hence his kinetic energy is 0.

then using conservation of energy principle:

potential energy of the spring+potential energy of the artist=total mechanical energy of the system

==>0.5*5*10^4*d^2-68*9.8*d=1932.9544

==>25000*d^2-666.4*d-1932.9544=0

solving for d , we get

d=0.291 m

Q2.

part A:

forces acting on the ski:

component of the weight along the incline, downwards=mass*9.8*sin(28)=4.6*mass N

component of the weight along perpendicular to the incline=mass*9.8*cos(28)=8.653*mass N

friction force=friction coefficient*normal force=0.09*mass*9.8*cos(28)=0.7787*mass N

hence along the incline, net force acting on thr ski=4.6*mass-0.7787*mass=3.8213*mass N

then acceleration=force/mass=3.8213 m/s^2

if final speed at the base of the incline is v m/s

initial speed=0 m/s

using the formula,

final speed^2-initial speed^2=2*acceleration*distance

==>v^2-0=2*3.8213*65

==>v=22.288 m/s

part B:

kinetic energy at the base of the incline=0.5*mass*v^2=248.377*mass J

as potential energy does not change while travelling across the snow, (as snow is level)

all the kinetic energy will be used as work against friction on the snow.

hence if distance covered before stopping is d ,

then friction force*d=248.377*mass

==>0.09*mass*9.8*d=248.377*mass

==>d=281.6 m

Q3.

let car A has mass 2*m and original speed Va.

car B has mass m and original speed Vb

as per the question,

kinetic energy of A=0.5*kinetic energy of B

==>0.5*2*m*Va^2=0.5*0.5*m*Vb^2

==>Vb^2=(1/0.25)*Va^2

==>Vb=2*Va...(1)

as per the question, if speed is increased by 12 m/s,

both have same kinetic energy

then 0.5*2*m*(Va+12)^2=0.5*m*(Vb+12)^2

using Vb=2*Va

(Va+12)^2=0.5*(2*Va+12)^2

==>2*(Va+12)^2=(2*Va+12)^2

==>2*(Va^2+2*Va*12+12^2)=4*Va^2+2*2*Va*12+12^2

==>2*Va^2+48*Va+288=4*Va^2+48*Va+144

==>2*Va^2=144

==>Va^2=72

==>Va=8.485 m/s

==>Vb=2*Va=16.97 m/s

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