The stretch of a spring constant k = 2000 N/m for its to store 210 j of energy;

Question

The stretch of a spring constant k = 2000 N/m for its to store 210 j of energy; A 1500 kg truck travelling north 31 km/hr turns east and accelerates to 41 km/hr or a) What is the change in the trucks kinetic energy in joules An electron with an initial velocity V_0 = 1.5 times 10^8 m/s enters a region of length L = 0.03 m where it is electrically accelerated it emerges with v = 5.70 times 10^9 m/s what is its acceleration, assumed constant A hoodlum throws a stone vertically downward with an initial speed of 12 ft/s from the roop of the building 30 ft above the ground. A) How long does it takes the stone to reach the ground A stone is projected at a cliff of height h with an initial speed of 25 m/s directed at an angle psi = 40 degree above the horizontal. The stone strikes at A = 3.5 sec after launching A) Find the height of the cliff, An elevator cab and it combined mass is 3200 kg. Find the tension in the supporting cable when the cab, originally moving downward at 20 m/s is brought to rest with a constant acceleration in a distance of 54 m

Explanation / Answer

Q8.

let stretch of the spring is x meters.

as energy stored =0.5*spring constant*stretch^2

==>210=0.5*2000*x^2

==>x=sqrt(210/(0.5*2000))=0.45826 m

Q9.

31 km/hr=31000/3600 m/s=8.6111 m/s

41 km/hr=41000/3600 m/s=11.389 m/s

then change in kinetic energy=0.5*mass*(final speed^2-initial speed^2)

=0.5*1600*(11.389^2-8.6111^2)

= 44446.622J

Q10.

initial speed=1.5*10^8 m/s

final speed=5.7*10^9 m/s

length=0.03 m

then acceleration=(final speed^2 - initial speed^2)/(2*distance)

=((5.7*10^9)^2-(1.5*10^8)^2)/(2*0.03)

=5.4113*10^20 m/s^2

Q11.

let time taken to reach ground be t seconds.

inital speed =12 ft/s

distance=30 ft

acceleration=32 ft/s^2

then 30=12*t+0.5*32*t^2

==>16*t^2+12*t-30=0

solving for t, we get
t=1.0447 seconds

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