Using the equation \\(U_{1} + K_{1} + W_{other} = U_{2} + K_{2}\\) , i.e, the Pr

Question

Using the equation (U_{1} + K_{1} + W_{other} = U_{2} + K_{2}) , i.e, the Principle of the Conservation of Energy, solve the following problem. A mass of 500 kg is attached over a massless pulley to a mass of 100g as shown below. Starting from rest at the top of the 45 degree incline shown, it is given a small tap (to break any residual forces of static friction), and moves under sliding friction ((mu_{k} = 0.2)) down the incline. After travelling 2.00 m, it encounters a spring (k = 45.0 N/m) in its equilibrium position. How fast is the half-kilogram mass moving once it has moved 220 cm from its starting position? Your answer should be accurate to 3 significant figures.

Explanation / Answer

the spring compressed by 0.2 m , energy stored in spring = 0.9 J

the small block has gone up by 2.2 m , gain in PE = 2.156 J

big block has gone down by 2.2 sin 45 , loss in PE = 7.622 J

friction has done work =1.524 J

change in kinetic energy = loss in potential energy - work done by friction - energy stored in spring

= 3.042 J

since both the blocks are moving with same speed

,0.5 V^2 (M+m) = 3.042

V= 3.184 m/sec

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