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Grade: 10
        A block of mass 5.0 kg is suspended from the end of a vertical spring which is stretched by 10 cm under the load of the block. The block is given a sharp impulse from below so that it acquires an upward speed of 2.0 m/s. How high will it rise? Take g = 10 m/s2.
5 years ago

Answers : (1)

Deepak Patra
askIITians Faculty
471 Points
							Sol. m = 5kg, x = 10cm = 0.1m, v = 2m/sec,
h =? G = 10m/sec^2
S0, k = mg/x = 50/0.1 = 500 N/m
Total energy just after the blow E = ½ mv^2 + ½ kx^2 …(i)
Total energy a a height h = ½ k (h - x)^2 + mgh …..(ii)
½ mv^2 + ½ kx^2 = ½ k (h - x)^2 + mgh
On, solving we can get,
H = 0.2 m = 20 cm

						
5 years ago
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