Question

Two blocks, of masses M = 2.3 kg and 2M are connected to a spring of spring constant k = 180 N/m that has one end fixed, as shown in the Figure-1. The coefficient of kinetic friction between the horizontal surface and the block is 0.12. The pulley is frictionless and has a negligible mass. The blocks are released from rest with the spring relaxed.

(a) What is the work done by the friction on the block of mass M?

(b) What is the combined kinetic energy of the two blocks when the hanging block has fallen 8 cm?

(c) What maximum distance does the hanging block fall before momentarily stopping?​

Answer 1

एथलेटिक्स का इतिहास और उसकी परिभाषा अथवास में आने वाली कठिनाइयों के बारे में बताइए

Answer 2

Answer:

a)  Work done by friction on the block of mass M = 1.326 J.

b) the combined kinetic energy of the two blocks when the hanging     block has fallen by 8 cm is 2.8832

c) The maximum distance the hanging block fall before it momentarily comes to rest is "x" = 48 cm

Explanation:

Given, M = 2.3 kg

k = 180 N/m

$\mu_k = 0.12$

a) By conservation of energy theorem,

decrease in potential energy of the block of 2M kg = increase in potential energy in spring + work done by friction. (As we are considering the situation when the blocks come to rest, so kinetic energy will be zero)

$2Mgx = \frac{1}{2} kx^2 + \mu_kMgx$  (where x = distance moved by the blocks).

$2Mg = \frac{1}{2} kx + \mu_kMg$

2(2.3)(10) = 0.5(180)(x) + 0.12(2.3)(10)  [using g value as 10m/s]

We get x = 0.48 m = 48 cm.

Therefore, work done by friction = $\mu_kMgx$ = 0.12(2.3)(10)(0.48) = 1.326 J.

b) Given, x = 8 cm = 0.08 m

Now by applying energy conversation,

   $2Mgx = \frac{1}{2} kx^2 + \mu_kMgx + K.E_M + K.E_{2M}$

2(2.3)(10)(0.08) = 0.5(180)(0.08)(0.08) + 0.12(2.3)(10)(0.08) + K.E(combined)

K.E(combined) =  3.68 - 0.576 - 0.2208

K.E(combined) = 2.8832

c) The maximum distance the hanging block fall before it momentarily comes to rest is "x" = 48 cm (As we can't use the equilibrium force conditions as we don't know about the coefficient of static friction, so we used the conservation of energy from (a)).

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