Chapter 8 Potential Energies A block of m=2.0kg is dropped from a height of 40cm

Question

Chapter 8 Potential Energies A block of m=2.0kg is dropped from a height of 40cm onto a spring with a k=1960 N/m. Find the maximum distance the spring will compress. What will the speed of the block be just when it reaches the spring? A spring with k= 170 N/m is at the top of a frictionless incline that has an angle of 37 degree . The end of the spring is lm from bottom of the incline when it is uncompressed. A 2.00kg canister is pushed against the spring until it is compressed 0.200m. (A) How much work was done to the canister at this point? (B) What is the speed of the canister when it leaves the spring at the equilibrium point? (C) If the surface now has a coefficient of friction of 0.35, what is the speed of the block when it reaches the bottom of the 1 m ramp? (D) How much thermal energy is lost? A 75g Frisbee is thrown from a point 1.1m above the ground with a speed of 12 m/s. When it reached a height of 2.1m, the speed is 10.5 m/s. What was the reduction of E_mcc of the Frisbee-Earth system because of air drag?

Explanation / Answer

Part 1:

From conservation of energy,

mgh = 1/2*k*x2

x = [2mgh/k]1/2

x = [2*2.0*9.81*0.4/1960]1/2

x = 0.8949 m

the maximum distance the spring will compress, x = 0.8949 m

v = [2gh]1/2

v = [2*9.81*0.4]1/2 = 2.8 m/s

the speed of the block when just it reaches the spring, v = 2.8 m/s

Part 2:

Mass = 75 g = 0.075 kg

Change in mechanical energy is : Emec = Eth

Emec = - [ K + U ]

Emec = 1/2 m (v12 - v22) + mg (h1 - h2)

Emec = 0.5*0.075*{( 12)2 - (10.5)2} + 0.075 * 9.8 * ( 1.1 - 2.1)

Emec = 0.53 J

The reduction in Emec of the Frisbee-Earth system because of the air-drag is 0.53 J

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