Physics Work, KE & PE Problem

[匿名]

2009-08-09 1:23 pm

1 ) Sam works at an amusement park. A boat of children on a water ride comes the the pier at the end of the ride. They are moving at a constant speed of 1.2 m/s, and the loaded boat has a mass of 1200kg. Sam slows the boat down by pushing on it until it comes to rest at the pier. If he stops it in a distance of 1.75m, and his arms make a 30o angle above the horizontal,

•a) how much work did Sam do on the boat?
•b) how large a force did he apply to the boat?

2 )A 50 kg block, initially at rest, falls from a height of 1.5 m onto a vertical spring that is fixed to the floor. The spring compresses 0.01 m. What is the spring constant of the spring?

3) A block of mass m slides down a frictionless ramp that becomes a circular loop of radius R. What must be the initial height h of the block above the ground for the centripetal force on the block at the top of the loop to be twice the weight of the block? (Notice that this is the centripetal force, not the net force.) Hint: The centripetal force on the block is given by Fc = mv2/R.

Is there anybody know how to solve the problems? im stuck.
Thxx

回答 (1)

Prof. Physics

2009-08-11 6:12 pm

✔ 最佳答案

1.a. Work done by Sam on the boat

= K.E. lost by the boat

= 1/2 mv2

= 1/2 (1200)(1.2)2

= 864 J

b. By work done = Fscos@

864 = F(1.75)cos30*

Force, F = 570 N

2. By the law of conservation of energy,

Loss of G.P.E. = Gain of E.P.E.

mg(h + x) = 1/2 kx2

(50)(10)(1.5 + 0.01) = 1/2 k(0.01)2

Spring constant, k = 1.51 X 107 Nm-1

3. At the top of the loop,

Centripetal force, Fc = mv2 / R

For centripetal force on top = 2 X weight

mv2 / R = 2mg

v2 = 2gR

By conservation of energy

Loss of G.P.E. = Gain of K.E.

mg(h - 2R) = 1/2 mv2

h - 2R = v2 / 2g

h - 2R = R

h = 3R

參考: Physics king

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