A block of mass m = 2 kg lies at rest next to a spring (k = 4300 N/m) on a flat

Question

A block of mass m = 2 kg lies at rest next to a spring (k = 4300 N/m) on a flat table top which has a coefficient of friction of 0.2. You push the block against the spring so that it compresses a distance of 2.2 cm. You release the block and it slides along the table until it comes to rest.
(a) What is the initial potential energy of the spring when it is compressed?
J

(b) What is the total work done by friction?
J

(c) From the point of release where the spring is compressed to the point where the block comes to rest, how far did the block move?
m

Explanation / Answer

a) Potential Energy stored in the spring = 1/2 k x^2
P.E = (1/2)*(4300)*(0.022)^2 = 1.0406 J
b)work done by friction during compression = (-µmg) * x =  (-0.2*2*9.8) *(.022) = -0.086 J

 workdone by friction after releasing the spring  = change in potential energy

= Final P.E - Initial P.E
= 0 - 1.0406 = -1.0406 J

=> Total work done =  -0.086 -1.0406 = -1.127 J
c) workdone by friction = frictional force * distance
-1.0406 = (-µmg) * d
=> d = 1.0406 / (0.2*2*9.8)
d = 0.265 mts

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