Question

A 2.50kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0350m . The spring has force constant 895N/m . The coefficient of kinetic friction between the floor and the block is 0.45 . The block and spring are released from rest and the block slides along the floor.


What is the speed of the block when it has moved a distance of 0.0150m from its initial position? (At this point the spring is compressed 0.0200m .)

Answer 1

Answer:

0.617m/s = 61.7cm/s

Explanation:

To find the values of the velocity you take into account the following formula:

$W_T=\Delta E_k=\frac{1}{2}m(v^2-v_o^2)$  (1)

That is, total work equals the change in the kinetic energy.

v_o: initial velocity = 0

m: mass = 2.50kg

The total work is:

$W_T=Fd+F_fd=kxd+(0.45)d$ (2)

x: the distance in wich the spring has been compressed

d: distance of the work

you will calculate v for a distance d = 0.0150m. By replacing in (2) and taking into account (2) you obtain:

$(895N/m)(0.0350m)(0.0150m)+(0.45N)(0.0150m)=0.476J\\\\v=\sqrt{\frac{2W_T}{m}}=\sqrt{\frac{2(0.476J)}{2.50kg}}=0.617\frac{m}{s}$

hence, the velocity is 0.617m/s = 61.7cm/s

Answer 2

The speed of the block is 0.404 m/s

Calculation of the speed:

Since a 2.50kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0350m . The spring has force constant 895N/m . The coefficient of kinetic friction between the floor and the block is 0.45

So,

$F_f=kmg=0.45*2.5*9.8=11.025N$

Now

$PE_i+W_{F_f}=PE_f+KE_f$

Now

$0.5kx_i^2-F_{f}*d=0.5kx_f^2+0.5mv^2$

Now

$0.5*895*0.035^2-11.025*0.015=0.5*895*0.02^2+0.5*2.5*v^2$

Therefore,

v = 0.404 m/s

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