Its question about velocity time graph?

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1 Answer
Apr 26, 2018

Magnitude of Frictional force acting on the block #=0.12\ N#.
We know that force of friction always opposes motion. As such when the block moves from #X# to #Y#. The frictional force produces a retardation. Which can be found from Newton's Second Law of Motion.

#a=-F/m=(-0.12)/0.15=-0.8\ ms^-2#

To find the velocity #v# of block when it hits #Y# can be found from the kinematic expression

#v=u+at# ........(1)

Inserting given values we get

#v=3+(-0.8xx2)#
#v=1.4\ ms^-1# ......(2)

The block rebounds from the wall and moves directly towards #X#
before coming to rest at the point #Z#.

Kinetic energy as it his the wall #=1/2mv^2=1/2xx0.15xx(1.4)^2#
#=0.147\ J#
Kinetic energy lost on hitting the wall#=0.072\ J#.
Remaining kinetic energy on rebound#=0.147-0.072=0.075\ J#
Velocity #v_r# as it leaves #Y# on rebound can be found from

#1/2mv_r^2=0.075#
#=>v_r=sqrt(2xx0.075/0.15)=+-1\ ms^-1#
Selecting negative root as now the velocity is in opposite direction of initial velocity.
#v_r=-1\ ms^-1# .......(3)

Time when block comes to rest at #Z# on rebound can be found from kinematic expression (1).
We note that after rebound magnitude of frictional force remains same but its direction changes. Hence we have

#0=-1+0.8t#
#=>t=1/0.8=1.25\ s# ......(4)

Once we have acceleration, velocity, and time we can sketch the velocity-time graph as shown below.

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