(4) The 4 kg block is dropped above the spring of constant 600 N/m and travels d
ID: 1756492 • Letter: #
Question
(4) The 4 kg block is dropped above the spring of constant 600 N/m and travels down without friction to strike the spring. If the velocity of the block immediately before it CE3302, Dynamics, Spring 2018 Midterm Exam #2 strikes the spring is 6 m/s, the maximum deformation of the spring after being struck by the spring is approximately: (a) 0.35 m (b) 0.43 nm c) 0.49 m 4 kg 0.56 m k 600 N/m (5) As shown, a block of 10 kg is moving to the left on the smooth surface at a velocity of 10 m/s when a force F, the magnitude of which varies with time as indicated, is applied in the direction shown. The velocity of the block at the moment the application of the force ends (f-8 s) is: (b) (c) (d) 4 m/s to the left 4 m/s to the right 16 m/s to the left 16 m/s to the right F (N) 10 0 0 6 8()Explanation / Answer
( 4 )
Kinetic Energy of the block before hitting the spring = ( 1 / 2 ) ( mass ) * ( velocity ) * ( velocity )
K.E. = 0.5 * 4 * 6 * 6 = 72 J
For maximum deformation of the spring, this Kinetic Energy will completely be transformed into potential energy of the spring, hence,
P.E. = 0.5 * k * x * x
where, x is the deformation
Now, 72 = 0.5 * 600 * x * x
On solving we get, x = 0.48989794855 m = 0.49 m, hence option C is correct.
( 5 )
Given is the variation of Force with time.
We know that, impulse is the area under this curve, which is also the change in moment of the body in same direction.
Area under the curve = 2 * [ 0.5 * 10 * 2 ] + [ 4 * 10 ] = 20 + 40 = 60 Ns
Note: Considering right side direction to be positive and let final velocity be v.
Impulse, 60 Ns = m ( final velocity - initial velocity ) = 10 * ( v - ( -10 ) ) = 10 * ( v +10 )
Hence, 60 = 10 * ( v + 10 ) i.e. ( v +10 ) = 6
Hence. v = -4 m/s
i.e. Moving with a velocity of 4m/s in left side. Hence, option A is correct.
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