a 3.00 kg block is accelerated from rest by compressed spring. The spring s init

Question

a 3.00 kg block is accelerated from rest by compressed spring. The spring s initial compression distance (delta x)=0.500m. the spring constant is 600N/m. The block leaves the springs relaxed length at A and then travels up frictionless 30" degree incline before coming to a stop at C.

b) calculate the total mechanical energy of the earth block spring system, assume the block has G potential energy at position A.

c) calculate the KE of the block at the instant that it leaves the spring.

d) use the conversation of mechanical energy to calculate the height , h to which the block slides along the incline before coming to a stop at C.

e) use mechanical energy conservation, calculate the speed of the block after it has slid 2.00m up the incline (position B)

Explanation / Answer

a) initial potential energy of the compressed spring = 0.5*k*x^2

= 0.5*600*0.5^2

= 75 J

b) total mechaincal energy = 75 J

c) Kinetic energy when it leaves the sprin = maximum potentaile enrgy of the spring

= 75 J

d) Apply, final mechanical energy = initial mechaincal enrgy

m*g*h = 75

h = 75/(m*g)

= 75/(3*9.8)

= 2.55 m

e) at this position, vertical height , h = 2*sin(30)

= 1 m

so, KE + PE = Total energy

KE + m*g*h = 75

KE = 75 - m*g*h

= 75 - 3*9.8*1

= 45.6 J

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