The 16-kg block A slides on the surface for which ,-0 3 The block has a velocity
ID: 1583926 • Letter: T
Question
The 16-kg block A slides on the surface for which ,-0 3 The block has a velocity u = 11 m/s when it is s 4 m from the 12-kg block B. The unstretched spring has a stiffness k 1200 N/m. Take e-0.6 The coefficient of friction is the same for both blocks Figure 1) Part A Determine the maximum compression of the spring due to the collision Express your answer to three significant figures and include the appropriate units. zmaz #10.594 11m Submit Previous Answers Request Answer X Incorrect: Try Again; 4 attempts remaining Provide Feedback Next Figure 1011 -kExplanation / Answer
We know the energy equations as
Kinetic energy = Ek = ½mv²
the spring energy is given by Esp = ½kx²
Work done is W = FS
the force of friction is Ff = N = mg
At s = 4m, the block A has a kinetic energy that will be reduced after the frictional force has worked against it during that 4 m distance.
thus the net energy at the moment of impact with block B will be the sum of the kinetic energy and the work done against the friction i.e,
Enet = Ek - Wf = ½mv² - mgs = m(½v² - gs)
This energy is transferred to the spring with 60% efficiency ( e=0.6)
thus
½kx² = 0.6m(½v² - gs)
solving this equation to find x, we get
x = [ (1.2m(½v² - gs)) / k ]
substituing the required values given , we get
x = [ (1.2*16*(½(11)² - 0.3*9.8*4)) / 1200 ] = 0.883 m (or 88.3 cm)
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.